# Full text of "The elements of Euclid, viz. the first six books, together with the eleventh and twelfth. The errors by which Theon, or others, have long ago vitiated these books are corrected, and some of Euclid's demonstrations are restored. Also, the book of Euclid's Data, in like manner corrected. By Robert Simson. 13th ed., carefully rev. and improved, to which is added a treatise on the construction of the trigonometrical canon; and a concise account of logarithms"

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Digitized by tine Internet Arciiive in 2007 witii funding from IVIicrosoft Corporation littp://www.arcli ive.org/details/elementsofeuclidOOeucluoft THE ELEMENTS OF EUCLID; VIZ, THE FIRST SIX BOOKS, TOGETHER WITH THE ELEVENTH AND TWELFTH. The ERRORS by which Theon, or others, have long ago vitiated these Books, are Corrected, And some of EUCLID'S DemoxNstrations are Restored. ALSO THE BOOK OF EU GLIDES DATA. In like Manner Corrected. % ROBERT SIMSON, M. D. Emeritus Professor of Mathematics in the University of Glasgow. The Thiiltiinth Edition, carefully revised and improyed, » To which is added, A TREATISE on the CONSTRUCTION of thk TRIGONOMETRICAL CANON i AND A CONCISE ACCOUNT of LOGARITHMS. LONDON: MINTED FOB r. WISGRAVE; J. JOHKSONJ W. J. AKD J. RICHARDSON; F. C. AND J. BIVIXGTON; J. walker; J. SCATCHERD AND C. J, tETTERMAN ; C. WILKIE AND J. ROBINSON; VERNOR, HOOD, Alfl> SHARPS ; LONGMAN, HURST, REES, AND OBME J T. CADIH AJTO W, DATIES; E. PHILUPS; AND J. MAWMAK. 1806. 90«184> WniGHi", Printn; Vo. W, St. John » S<|»ai*, Clertf»W«U. ^ % TO THE KING, THIS EDITION «r TKX PRINCIPAL BOOKS Ot TU ELEMENTS OF EUCLID, AX9 or TSX BOOK OF HIS DATA, IS MOST BVMBLT DSDICATSD, BT HIS MAJESTY'S MOST DUTIFUL, AND MOST DBVOTBD SUBJECT AND SERVANT, ROBERT SIMSON. CONTENTS. PREFACE, The ELEMENTS, * Page i to 2^ NOTES, Critical and Geometrical, -- 287. The DATA, 355 Plane TRIGONOMETRY, 473 CONSTRUCTIONSof the TRIGONOMETRICAL CANON, r 48s Of LOGARITHMS, 49^ Spherical TRIGONOMETRY, 504 PREFACE. Jl HE opinions of the moderns concerning the author of the Elements of Geometry, which go under Euclid's name, are very different and contrar}' to one another. Peter Ramus as- cribes the Propositions, as well as their Demonstrations, to Theon ; others think the Propositions to be Euclid's, but that the Demonstrations are Theon's ; and others maintain, that all the Propositions and their Demonstrations are Euclid's own. John Buteo and Sir Henry Savile are the authors of greatest note who assert this last ; and the greater part of geo- meters have ever since been of this opinion, as they thought it the most probable. Sir Henry Savile, after the several argu- ments he brings to prove it, makes this conclusion (Page 13. Praeleft.) '* That, excepting a very few interpolations, expli- " cations, and additions, Theon altered nothing in Euclid." But, by often considering and comparing together the Defi- nitions and Demonstrations as they are in the Greek editions we now have, I found that Theon, or whoever was the editor of the present Greek text, by adding some things, suppressing others, and mixing his own with Euclid's Demonstrations, had changed more things to the worse than is commonly sup- posed, an<l those not of small moment, especially in the fifth and eleventh Books of the Elements, which this editor has greatly vitiated ; for instance, by substituting a shorter, but insufficient Demonstration of the iSthProp. ofthe 5th Book, in place of the legitinwite one which Euclid had given ; and by taking out of this Book, besides other things, the good defi- nition which Eudoxus or Euclid had given of compound ratio, and giving an absurd one in place of it in the 5th Defini- tion of the 6th Book, which neither Euclid, Archimedes, Apellotiius, nor any geometer before Theon's time, ever made usp of, and of which there is not to be found the least appearance in any of their writings j and, as this Definition did much embarrass beginners, and is quite useless, it is now thrown out of the Elements, and another, which, without <ioubt, Euclid had given, is put in its proper place among the Definitioo^ VI PREFACE. Definitions of the 5th Book, by which the doctrine of com- pound ratios is rendered plain and easy. Besides among the Definitions of the i ith Book, there is this, which is the tenth, viz. ** Equal and similar solid figures are those which arc ** contained by similar planes of the same number and mag- " nitude." Now this Proposition is a Theorem, not a Defi- nition ; because the equality of figures of any kind must be demonstrated, and not assumed ; and therefore, though this were a triie Proposition, it ought to have been demonstrated. But, indeed, this Proposition, which makes the loth Defini- tion of the nth Book, is not true universally, except in the case in which each of the solid angles of the figures is con- tained by no more than three plane angles j for in other cases, two solid figures may be contained by similar planes of the same number and magnitude, and yet be unequal to one ano- ther, as shall be made evident in the Notes subjoined to these Elements. In like manner, in the Demonstration of the 26th Prop, of the i ith Book, it is taken for granted, that those solid angles are equal to one another which are contained by plain angles of the same number and magnitude, placed ia the same order; but neither is this universally true, except in the case in which the solid angles are contained by no more than three plane angles ; nor of this case is there any Demon- stration in the Elements we now have, though it be quite ne- cessary there should be one. Now, upon the ioth Definition of this Book depend the 25th and 28th Propositions of it ; and, upon the 25th and 26th depend other eight, viz* the 27th, 31st, 32d, 33d, 34th, 36th, 37th, and 40th of the same Book; and the 12th of the 1 2th Book depends upon the eighth of the same; and this eighth, and the Corollary of Proposition 17th and Proposition i8th of the I2th Book, depend upon the gth D finition of the nth Book, which is not a right defini- tion ; because there may be solids contained by the same num- ber of similar plane figures, which are not similar to one ano- ther, in the true sense of similarity received by geometers ; and all these Propositions have, for these reasons, been insuf- ficiently demonstrated since Theon's time hitherto. Besides, there are several other things, which have nothing of Euclid's accuracy, and which plainly shew, that his Elements have been much corrupted by unskilful geometers; and though these are not so gross as the others now mentioned, they ought by no means to remain uncorrected. Upon these accounts it appeared necessary, and I hope will prove acceptable, to all lovers of accurate reasoning, and of PRtFACE. VU mathematical learning, to remove such blemishes, and restore the principal Books of the Elements to their original accuracy, as far as I was able ; especially since these Elements are the foundation of a science by which the investigation and disco- very of useful truths, at least in mathematical learning, is pro- moted as far as the limited powers of the mind allow j and which likewise is of the greatest use in the arts both of peace and war, to many of which geometry is absolutely necessary. This I have endeavoured to do, by taking away the inaccu- rate and false reasonings which unskilful editors have put into the place of some of the genuine Demonstrations of Euclid, who has ever been justly celebrated as the most accurate of geometers, and by restoring to him those things which Theon or others have suppressed, and which have these many ages been buried in oblivion. In this edition, Ptolemy's Proposition concerning a pro- perty of quadrilateral figures in a circle, is added at the end of the sixth Book, Also the Note on the 29th Proposition, Book 1st, is altered, and made more explicit, and a more gene- ral Demonstration is given, instead of that which was in the Note on the lOth Definition of Book nth; besides,, the Translation is much amended by the friendly assistance of a learned gentleman. To which are also added, the Elements of Plane and Sphe- fical Trigonometry, which are commonly taught after the Elements of Euclid. ADVERTISEMENT. J- HE favourable reception which former edi- tions of Professor Simson^s Elements of Euclid have met with from the public^ induced the pro- prietors of the work to carry into execution every measure most likely to secure and continue general approbation. With this view, the pre- sent edition has been carefully revised throughout, by a very eminent mathematician; for the conve- nience of tutors, as well as students, a short treatise on the Construction" of the Trigo- nometrical Canon has now been inserted, from a late celebrated author ; and to this has been added, a co?icise Account o/" Logarithms, and improved methods of calculating them, by the present Savilian Professor of Geometry in the Univej'sity of Oxfords THE THE ELEMENTS OF EUCLID. BOOK I. DEFINITIONS. I. A POINT is that which hath no parts, or which hath no book i. magnitude. v^^v,^^./ J{^ See Notes. A line is length without breadth. III. The extremities of a line are points. IV. A straight line is that which lies evenly between its extreme points. V. A superficies is that which hath only length and breadth. VI. The extremities of a superficies are lines. VII. A plane superficies is that in which any two points being taken, See N. the straight line between them lies wholly in chat superficies. VIII. ** A plane angle is the inclination of two lines to one another See N, " in a plane, which meet together, but are not in the same *' direction." IX. A plane redlilineal angle is the inclination of two straight lines to one another, which meet together, but are not in the same straight line. B N.B. Book I. THE ELEMENTS N. B. ' When several angles are at one point B, any one « of them is expressed by three letters, of which the letter that * is at the vertex of the angle, that is, at the point in which * the straight lines that contain the angle meet one another, ' is put between the other two letters, and one of these two is ' somewhere upon one of those straight lines, and the other ' upon the other line : Thus the angle which is contained by * the straight lines AB,CB, is named the angle ABC, or CB A; * that which is contained by AB, BD is named the angle * ABD, or DBA ; and that which is contained by BD, CB ' IS caUed the angle DBC, or CBD ; but, if there be only ' one angle at a point, it may be expressed by a letter placed ' at that point ; as the angle at E.' When a straight line standing on ano- ther straight line makes the adjacent angles equal -to one another, each of the angles is called a right angle ; and the straight line which stands on the other is called a perpendicular to it. XI. An obtuse angle is that which is greater than a right angle. Xli. An acute angle is that which is less than a right angle. XIII. " A term or boundary is the extremity of any thing." A figure is that which is inclosed by one or more boundaries. OF EUCLID. 3 XV. ^^^ A circle is a plane figure contained by one line, which is cal- ^^^^"""^ led the circumference, and is such that all straight lines drawn from a certain point within the figure to the cir- cumference, are equal to one another. (C XVI. And this point is called the centre of the circle. XVII. A diameter of a circle is a straight line drawn through the See N. centre, and terminated both ways by the circumference. xvm: A semicircle is the figure contained by a diameter and the part of the circumference cut ofF by the diameter. XIX. A segment of a circle is the figure contained by a straight " line, and the circumference it cuts off." XX. Reftilineal figures are those which are contained by straight lines. XXI. Trilateral figures, or triangles, by three straight lines. XXII. * Quadrilateral, by four straight lines. XXIII. Mutilateral figures, or polygons, by more than four straight lines. XXIV. Of three-sided figures, an equilateral triangle is that which has three equal sides. XXV. An isosceles triangle is that which has only two sides equal. B2 THE ELEMENTS Book T. XXVI. A scalene triangle, is that which, has three unequal sides. XXVII. A right angled triangle, is that which has a right angle. XXVIII. An obtuse angled triangle, is that which has an obtuse angle. XXIX. An acute angled triangle, is that which has threeacu^e angles. XXX. Of four-fided figures, a square is that which has all its sides equal, and all its angles right angles. XXXI. An oblong, is that which has all its angles right angles, but has not all its sides equal. XXXII. A rhombus, is that which has its sides equal, but its angles are not right angles. XXXIII. iSce N. A rhomboid, is that which has its opposite sides equal to one another, but all its sides are not equal, nor its angles right angles. OF EUCLID. 5 XXXIV. ^2^ All other four-sided figures besides these, are called Trape- ^^^^^^ ziums. XXXV. Parallel straight lines, are such as are in the same plane, and which, being produced ever so far both ways, do not meet. POSTULATES. T ^ X-iET it be granted that a straight line may be drawn from any one point to any other point. II. That a terminated straight line may be produced to any length in a straight line. ^ III. And that a circle may be described from any centre, at any distance from that centre. AXIOMS. rp I. J. HINGS which are equal to the same are equal to one another. IL ^ - - If equals be added to equals, the wholes are equal, '^-i- "- = -^ III. 'If equals be taken from equals, the remainders are equal. IV. If equals be added to unequals, the wholes are unequal. V. If equals be taken from unequals, the remainders are unequal. VI. Things which are double of the same, are equal to one another. VII. Things which are halves of the same, are equal to one another. VIII. Magnitudes which coincide with one another, that is, which exadly fill the sams space, are equal to one another. B3 ) THE ELEMENTS Book I. IX. *'"*'''^"*''^ The whole is greater than its part. X. Two straight lines cannot incfose a space. XI. All right angles are equal to one another. XII. *' If a straight line meets two straight lines, so as t© make the two interior angles on the same side of it taken toge- ther less than two right angles, these straight lines being continually produced, shall at length meet upon that side " on which are the angles which are less than two right '' angles. See the notes on Prop. 29. of Book I." cc OF EUCLID. PROPOSITION I. PROBLEM. A O describe an equilateral triangle upon a given iinite straight line. Let AB be the given straight line; it is required to describe an equilatecal triangle uf>on it. From the centre A, at the dis- tance AB, describe* the circle BCD, and from the centre B,at the distance B/\, describe the' circle ACE ; and from the point D C, in which the circhs cut one another, draw the straight lines'' CAjCB to the points A,B; ABC shall be an equilateral triangle. Because the point A is the centre of the circle BCD, AC is equal "^to AB ; and. because the point B is the centre of the circle ACt, BC is equal to BA : But it has been proved that CA is equal to AB ; therefore CA,CB are each of them equal to AB ; but things which are equal to the same are equal to one another ^ ; therefore CA is equal to CB ; wherefore CA, AB, BC are equal to one another ; and the triangle ABC is therefore equilateral, and it is described upon the given straight line AB. Which was required to be done. PROP. II. PROB. £ ROM a given point to draw a straight line.equal to a o-iven straight line. o o Let A be the given point, and BC the given straight line ; it is required to draw from the point A a straight line equal toBC From the point A to B draw» the straight line AB ; and upon it de- scribe'' the equilateral triangle DAB, and produce"^ the straight lines Da, DB, to E and F ; from the centre B, at the distance BC, describe** the circle CGH, and from the centre D, at the distance DG, describe the circle GK!L. AL shall be equal to BC. B 4 Because Book I. Postu- late. "IPost <= 15 Definj- tiOD. " lit Axi- om. M Post. 1. «2Po»t. 3Fot 8 THE ELEMENTS Bgok r. Because the point B is the centre of the circle CGH, BC is •^TdT^ ^^li^l^ ^o BG ; and because Dis the centre of the circle GKL, DL is equal to DG, and DA, DB, parts of them, are equal ; '3 Ax. therefore the remainder AL is equal to the remainder*^ BG: But it has been shown, that BC is equal to BG j wherefore AL and BC are each of them equal to BG j and things that are equal to the same are equal to one another j therefore the straight line AL is equal to BC. Wherefore from the given point A a straight line AL has been drawn equal to the given straight line BC. Which was to be done. PROP. 111. PROB. F] ROM the greater of two given straight lines to cut off a part equal to the less. Let AB and C be the two given straight lines, whereof AB is the greater. It is required to cut off from AB, the greater, a part equal to C, the less. •2.1. From the point A draw* the straight line AD equal to C ; and from the centre A, and at the dis- "»3Post. tance AD, describe'' the circle DEF ; and because A is the centre of the circle DEFj AE shall be equal to AD ; but the straight line C is likewise equal to AD ; whence AE and C are each of them equal to AD ; wherefore the straight line AE is equal •= 1 Ax. to '^C, and from AB, the greater of two straight lines, a part AE has been cut oft equal to C the less. Which was to be done. PROP. IV. THEOREM. J F two triangles have two sides of the one equal to two sides of the other, each to each ; and have like- wise the angles contained by those sides equal to one anotlier; they shall likewise have their bases, ov third sides, equal ; and the two triangles shall be equal ;' and their other angles shall be e(jual, each to each, viz. those to which the equal sides are opposite. Let ABC, DEF be two triangles, which have the two sides AB, AC equal to the two sides DE, DF, each to each, viz. 2 AB OF EUCLID. Book. I. ABtoDE,andACtoDF; and the angle BAG equal to the angle EDP\ the base BC shall be equal to the base EF ; and the triangle ABCto the triangle DEF; and the other angles to which the equal sides are opposite,shall be equal each to each, viz. the angle ABC to the angle DEP\ and the angle ACB to DFE. For, if the triangle ABC be applied to DEF, so that the point A may be on D, and the straight line A3 upon DE ; the point B shall coincide with the point E, because AB is equal toDE ; and AB coinciding with DE, AC shall coincide with DF, because the angle BAG is equal to the angle EDF ; wherefore also the point C shall coincide with the point F, because the straight line AC is equal to DF : But the point B coincides with the point E ; wherefore the base BC shall coin- cide with the base EF, because the point B coinciding with E, and C with F, if the base BC does not coincide with the base EF, two straight lines would inclose a space, which is impossi- blc\ Therefore the base BC shall coincide with the base EF. « 10 Ax. and be equal to it. Wherefore the whole triangle ABC shall coincide with the whole triangle DEF, and he equal to it ; and the other angles of the one shall coincide with the remaining angles of the other, and be equal to them, viz. the angle ABC to the angle DEF, and the angle ACB to DFE. Theretore, if two triangles have two sides of the one equal to two sides of the other, each to each, and have likewise the angles con- tained by those sides equal to one another, their bases shall likewise be equal, and the triangles be equal, and their other angles to which the equal sides are opposite shall be equal, each to each. Which was to be demonstrated. • PROP. V. THEOR. 1 HE angles at the base of. an Isosceles triangle are equal to one another ; and if the equal sides be produced, the angles upon the other side of the base shall be equal. Let ABC be an Isosceles triangle, of which the side AB is { equal Book I. 3.1. '>4. 1. I Ax. THE ELEMENTS equal to AC, and let the straight lines AB, AC be produced to D and E, the angle ABC shall be equal to the angle ACB, and the angle CBD to the angle BCE. In BD take any point F, and from AE the greater, cut off AG equal* to AF, the less, and join FC, GB. Because AF is equal to AG, and AB to AC, the two sides FA, AC are equal to the two GA, AB, each to each j and they contain the angle FAG common to the two triangles AFC, AGB ; therefore the base FC is equal '' to the base GB, and the triangle AFC^ to the triangle AGB j and the remain- ing angles of the one are equaP to the remaining angles of the other,each to each, to which the equal sides are opposite ; viz. the angle ACF to the angle ABG, and the angle AFC to the angle AGB : And because the whole AF, is equal to the whole, AG of which the parts AB, AC, are-L' equal ; the remainder BF shall be equal '^ to the remainder CG ; and FC was proved to be equal to GB ; therefore the two sides BF, FC are equal to the two CG, GB, each to each ; and the angle BFC is equal to the angle CGB, and the base BC is common to the two trian- gles BFC, CGB J wherefore the triangles are equal'', and their remaining angles, each to each, to which the equal sides are opposite ; therefore the angle FBC is equal to the angle GCB, and the angle BCF to the angle CBG : And, since it has been demonstrated, that the whole ^ngle ABG is equal to the whole ACF, the parts of which, the angles CBG, BCF are also equal 5 the remaining angle ABC is therefore equal to the remaining angle ACB, which are the angles at the base of the triangle ABC: And it has also been proved that the angle FBC is equal to the angle ^CB, which are the angles upon the other side of the base. Therefore the "angles at the base. Sec. Q^E.D. Corollary. Hence every equilateral triangle is also equiangular. PROP. VI. THEOR. , X F two angles of a tjj^nglc be eijual to one ano- ther, the sides also which subtend, or arc opposite tOy the equal angles, shall be equal to one another. Let b-i. 1. OF EUCLID. II Let ABC be a triangle having the angle ABC equal to the ^°°^ ^• angle ACB ; the side AB is also equal to the side AC. ^-^/^«' For, if AB be not equal to AC, one of them is greater than the other : Let AB be the greater ; and from it cut* ofF DB » 3- 1. equal to AC, the less, and join DC ; there- fore, because in the triangles DBC, ACB, DB is equal to AC, and BC common to both the two sides, DB, BC are equal to the two AC, CB each to each ; and the angle DBC is equal to the angle ACB ; therefore the base DC is equal to the base AB, and the triangle DBC is equal to the triangle'' ACB, the less to the greater ; which is absurd. Therefore AB is not L" unequal to AC, that is, it is equal to it. Wherefore, if two angles, &c. Q. E. D. Cor. Hence every equiangular triangle is also equilateral. PROP. Vn. THEOR. L PON the same base, and on the same side of it, s** n, there cannot be two triangles that have their sides which are terminated in one extremity of the base equal to one another, and likewise those which are terminated in .tlie other extremity. If it be possible, let there be two triangles ACB, ADB, up- on the same base AB, and upon the same side of it, which have their sides CA, DA, terminated in the extremity A of the base equal to one another, and liice- wise their sides CB, DB,that are ter- minated in B. Join CD ; then, in the case in which the vertex o^ each of the tri- angles is without the other triangle, because AC is equal to AD, the , ^ -. , ,^ ^ angle ACD is equal* to the angle ADC : But the an^le ACD is greater than the angle BCD; therefore the angle ADC is greater also than BCD; much more then is the angle BDC greater than the angle BCD. Again, because CB is equal to DB, the angle BDC is equal* to the angle BCD ; but it has been demonstrated to be greater than it j which is impossible. But 12 THE ELEMENTS 3. J. Book 1. gut if one of the vertices, as D, be within the other triangle ^'^^^'"^ ACB ; produce AC, AD to E, F •, therefore, because AC is equal to AD in the triangle ACD, the angles ECD, FDC upon the other side of the bas'i CD are equal^ to one another, but the angle ECD is greater than the angle BCD ; wherefore the angle FDC is likewise greater than BCD ; much more then is the angle BDC greater than the angle BCD. Again, because CB is equal to DB, the angle BDC is equal "^ to the angle BCD j but BDC has been proved to be greater than the same BCD ; which is unpossible. The case in which the vertex of one triangle is upon a side of the other, needs no demonstration. Therefore, upon the same base, and on the same side of it, there cannot be two triangles that have their sides which are terminated in one extremity of the base equal to one another, and likewise those which are terminated in the other extre- mity. Q.E.D. PROP. VIII. THEOR. If two triangles have two sides of the one equal to two sides of the other, each to each, and have like- wise their bases equal ; the angle which is contain- ed by the two sides of the one shall be equal to the angle contained by the two sides equal to them, of the other. Let ABC, DEF be two triangles, having the two sides AB, AC, equal to the two sides DE, DF, each to each, viz. AB to DE, and AC a ^^ P to DF ; and also the baseBC equal to the base EF. The angle BAG is equal to the angle EDF. For, if the tri- angle ABC be applied to DEF, so that the point B be on E, and the straight line BC upon EF ; the point C shall also coincide with the point F. Because BC OF EUCLID. ^3 BC is equal to EF ; therefore BC coinciding with EF ; BA BookL and AC shall coincide with ED and DF ; for, it" the base BC ^""^'^■^ coincides with the base EF, but the sides BA, CA do not coincide with the sides ED, FD, but have a different situation as EG, FG, then, upon the same base EF, and upon the same side of" it, there can be two triangles that have their sides which are terminated in one extremity ot the base equal to one ano- ther, and likewise their sides terminated in the other extremity : But this is impossible* ; therefore, if the base BC coincides* '• '• with the base EF, the sides BA, AC cannot but coincide with the sides ED, DF ; wherefore likewise the angle BAC coin- cides with the angle EDF, and is equal *• to it. Therefore if ^ ^^ two triangles, &c. Q^E, D. PROP. IX. PROB. A O bisect a given rectilineal angle, that is, to di- vide it into two equal angles. Let BAC be the given rectilineal angle, it is required to bisect it. Take any point D in AB, and from AC cut ' off AE equal • 3- i- to AD; join DE, and upon it describe** •> 1. 1. an equilateral tiiangle DEF; then join AF ; the straight line AF blsedls the angle BAC. Because AD is equal to AE, and AF is common to the two triangles DAF, EAFj the two sides DA, AF, are equal to the tWo sides EA, AF, each to each ; and the < base DF is equal to the base EF ; there--! fore the angle DAF is equal<^ to the angle EAF ; wherefore the given rectilineal angle BAC is bisefted by the straight line AF. Which was to be done. PROP. X. PROB. •=8. 1. 1 O bisect a given finite straight Hne, that is, to divide it into two equal parts. Let AB be the given straight line ; it is required to divide 4t into two equal parts. Describe* upon it an equilateral triangle ABC, and bise<Sl * ^ ''the angle ACB by the straight line CD. AB is cut into two equal parts in the point D. Because 1. 9. 1. Book I. ■4,1. THE ELEMENTS Because AC is equal to CB, and CD common to the two triangles ACD, BCD ; the two sides AC, CD are equal to BCj CD, each to each ; and the angle ACD is equal to the angle BCD ; therefore the base AD is equal to the base'^ DB, and the straight line AB is divided into two equal parts in the point V D. Which was to be done. SeeN. » .3. 1 . »1. 1. «8. 1. <»10Def. PROP. XI. PROB. X O draw a straight line at right angles to a given straight line, from a given point in the same. • Let AB be a given straight line, and C a point given in it ; it is required to draw a straight line from the point C at right angles to AB. Take any point D in AC, and ^ make CE equal to CD, and upon DE describe ^ the equi- lateral triangle DFE, and join FC, the straight line FC drawn from the given point C is at right angles to the given straight line AB.. Because DC is equal to CE, and FC common to the two ^ triangles DCF,ECFj the two sides DC, CF, are equal to the two EC,CF, each to each ; and~ the base DF is equal to the base EF j therefore the angle DCF is equal "^ to the angle ECF ; and they are adjacent angles. But, when the adjacent angles which one straight line makes with another straight line are equal to one anothcr,each of them is called a right'' angle j therefore each of the angles DCF, ECF, is a right angle. Wherefore, from the given point C, in the given straight line AB, FC has been drawn at right angles to AB. Which was to be done. Cor. By help of this problem, it may be demonstrated, that two straight lines cannot have a common segment. If it be possible, let the two straight lines ABC, ABD have the segment AB common to both of them. Fiom the point B draw BE at right angles to AB j and because ABC is a straight line, OF EUCLID. line, the angle CBE is equal * to the angle EBA ; in the same manner, because ABD is a straight line, the angle DBE is equal to the angle EBA ; where- fore the angle DBE is equal to the angle CBE, the less to the greater ; which is impossible ; therefore two straight lines can- not have a common segment. E A D PROP. XII. PROB. X O draw a straight line perpendicular to a given straight line of an unlimited length, from a given point witiiout it. Let AB be the given straight line, which may be produced to any length both ways, and let C be a point without it. It is required to draw a straight line perpendicular to AB from the point C, Take any point D upon the other side of AB, and from the centre C, at the distance CD, describe'' the circle EGF meet- ing AB in FG ; and bisect "^FG in H, and join CF, CH, CG ; the straight line CH, drav/n from the given point C, is per- pendicular to the given straight line AB. Because FH is equal to HG, and HC common to the two triangles FHC, GHC, the two sides FH, HC are equal to the two GH, HC, each to each ; and the baseCF is equaH to the « 15 Dcf. base CG ; therefore the angle CKF is equal • to the angle j CHG ; and they are adjacent angles j but when a straight line standing on a straight line makes the adjacent angles equal to one another, each of them is a right angle ; and the straight line which stands upon the other is calbd a perpendicular to it ; therefore from the given point C a perpendicular CH has been drawn to the giYen straight line AB. Which was. to be done. PROP. XIII. THEOR. X HE angles which one straight line makes with another upon the one side of it, are either two right angles, or are together equal to two right angles. Let s, 1. • Def. 10. «>11. 1. «2Ax. ••l Ax. THE ELExMENTS Let the straight line AB make with CD, upon one side of it, the angles CBA, ABD ; these are either two right angles, or are together equal to two right angles. For if the angle CBA be equal to ABD, each of them is a E A D A B right ^ angle ; but, if not, from the point B draw BE at right angles ^ to CD ; therefore the angles CBE, EBD are two right angles* ; and because CBE is equal to the two angles CBA; ABE together, add the angle EBD to each of these equals j therefore the angles CBE, EBD are equal "^ to the three an- gles CBA, ABE, EBD. Again, because the angle DBA is equal to the two angles DBE, EBA, add to these equals the angle ABC, therefore the angles DBA, ABC are equal to the three angles DBE, EBA, ABC ; but the angles CBE, EBD have been demonstrated to be equal to the same three angles; and things that are equal to the same are equal** to one another ; therefore the angle CBE, EBD are equal to the angles DBA, ABC ; but CBE, EBD are two right angles ; therefore DBA, ABC are together equal to two right angles. Wherefore when a straight line, &c. Q. E. D. PROP. XIV. THEOR. XF, at a point in a straight line, two other straight lines, upon the opposite sides of it, make the adja- cent angles, together equal to two right angles^ these two straight lines shall be in one and the same straight line. ' At the point B in th:; straight .\ line AB,let the two straight lines BC, BD upon the opposite sides of AB, make the adjacent an- gles ABC, ABD equal together to two right angles, BD is in the same straight line with CB. For, ifBDbenot in the sameC- straight line with CB, let BE be B D in OF EUCLID. 17 in the same straight line with it ; therefore, because the straight BookL line AB makes angles with the straight line CBE, upon one side of it, the angles ABC, ABE are together equal* to two »13. 1. right angles; but the angles ABC, ABD are li ice wise toge- ther equal to two right angles ; therefore the angles CBA, ABE are equal to the angles CBA, ABD : Take away the common angle AEC, the remaining angle ABE is equaP to the re- "3. As. maining angle ABD, the less to the greater, which is impos- sible ; therefore BE is not in the same straight line with BC. And, in like manner, it may be demonstrated, that no other can be in the same straigh: line with it but BD, which there- fore is in the same straight line with CB. Wherefore, if at a point, &c. Q^ E. D. PROP. XV. THEOR. 1 F two straight lines cut one another, the vertical, or opposite, angles shall be equal. Let the two straight lines AB, CD, cut one another in the point E ; the angle AEC shall be equal to the angle DEB, and CEB to AED. Because the straight line AE makes with CD the angles CE A, ^ AED, these angles are together equal * to two right angles. — Again, because the straight line DE makes with AB the angles AED, DEB, these also are to- gether equal* to two right an- gles; and CEA, AED, have been demonstrated to be equal to two right angles ; wherefore the angles CEA, AED, are equal to the angles AED, DEB. Take away the common angle AED, and the remaining angle CEA is equal'' to the remaining angle DEB. In the same manner "3- -Ax, it can be demonstrated, that the angles CEB, AED are equal. Therefore, if two straight lines, &c. Q. E. D. Cor. I. From this it is manifest, that, if two straiglit lines cut one another, the angles they make at the point where they cut, are together equal to four right angles. Cor. 2. And consequently that all the angles made by any number of lines meeting in one point, are together equal to four right angles. C M3. I. iS THE ELEMENTS Book I. »10. 1. *lb. 1. *. 1. 'IS. 1. PROP. XVI. THEOR. If one side of a triangle be produced, the exterior angle is greater than either of the interior opposite angles. Let ABC be a triangle, and let its side BC be produced to D, the exterior angle ACD is greater than either of the inte- rior opposite angles CBA, BAC. Bisect* AC in E, join BE A and produce it to F, and make EF equal to BE j join also FC, and produce AC to G. Because AE is equal to EC, and BE to EF ; AE, EB are equal to CE, EF, each to each j and the angle H AEB, is equal** to the angle CEF, because they are op- posite vertical angles ; there- . fore the base AB is equa^ to the base CF, and the triangle AEB to the triangle CEF, and the remaining angles to the remaining angles, each to each, to which the equal sides are opposite i wherefore the angle BAE is equal to the angle ECF i but the aiigle ECD is greater than the angle ECF ; therefore the angle ACD is greater than BAE : In the same manner, if the side BC be bisected, it may be demonstrated that the angle BCG, that is"*, the angle ACD, is greater than the angle ABC. Therefore, if one side, &c. Q; E. D. '\6. 1. A PROP. XVIL THEOR. NY two angles of a triangle are together less than two rioht an":les. Let ABC be any triangle ; any two of its angles together are less than two right angles. Produce BC to Dj and be- cause ACD is the exterior ansle of the triangle ABC, ACDTis greater' than the interior and oppo^te angle ABC ; to each of OF EUCLID. '9 •3. 1. " 1&. 1. these add the angle ACB ; therefore the angles ACD, ACB Boor I. are greather than the angles ABC, ACB ; but ACD, ACB '"^^''^^ are together equal** to two right angles; therefore the angles » 13. 1. ABC, BCA are less than two right angles. In like manner, it may be demonstrated, that BAC, ACB, as also CAB, ABC, are less than two right angles. Therefore any two angles, &c. Q, E. D. PROP. XVIII. THEOR. JL HE greater side of every triangle is opposite to the greater angle. Let ABC be a triangle, of which the side AC is greater than the side AB ; the angle ABC is also greater than the angle BCA. Because AC is greater than AB, make » AD equal to AB, and join BD j and because ADB is the exterior angle of the tri- angle BDC, it is greater ^ than the interior and opposite angle DCB j but ADB is equal^ to «5. 1. ABD, because the side AB is equal to the side AD ; therefore the angle ABD is likewise greater than the angle ACB. Wherefore much more is the angle ABC greater than ACB. Therefore the greater side, &c. Q^ E. D. PROP. XIX. THEOR. X HE greater angle of every triangle is subtended by the greater side, or has the greater side opposite to it. Let ABC be a triangle, of which the angle ABC is greater than the angle BCA ; the side AC is likewise greater than the side AB. For, if it be not greater, AC, must either be equal to AB, or less than it j it is not equal, be- cause then the angle ABC would be equal * to the angle ACB ; but It is not J therefore AC is not equal to AB j neither is it less ; became ikca the. angle _ Cl ABC ^. I. 20 THE ELEMENTS Book i,^ ABC would be less '' than the angle ACB i but it is not i " ~ ' therefore the side AC is not less than AB ; and it has been shewn that it is not equal to AB ; therefore AC is greater than AB. Wherefore the greater angle, &c. (^ E. D. " 18. 1. PROP. XX. THEOR. SeeN. Any two sidcs of a triangle are together greater than the third side. Let ABC be a triangle j any two sides of it together are greater than the third side, viz. the sides BA, AC greater than the side BC ; and AB, BC greater than AC ; and BC, CA greater than AB. Produce BA to the point D, •3. J- and make^ AD equal to AC j and join DC. Because DA is equal to AC, the angle ADC is likewise equal »- 5. 1. b to ACD ; but the angle BCD is greater than the angle ACD ; therefore theangle BCD is great- er than the angle ADC ; and because the angle BCD of the triangle DCB is greater than its angle BDC, and that the « 19 1. greater*^ side is opposite to the greater angle; therefore the side DB is greater than the side BC ; but DB is equal to BA and AC ; therefore the sides BA, AC are greater than BC. In the same manner it may be demonstrated, that the sides AB, BC are greater than CA, and BC, CA greater than AB. Therefore any two sides, &c. Q^ E. D. PROP. XXL THEOR. 5eeN. If, from, the ends of the side of a triangle, there be drawn two straight lines to a point within the trian- gle, these shall be less than the other two sides of the triangle, but shall contain a greater angle. Let the two straight lines BD, CD be drawn from B, C, the ends of the side BC of the triangle ABC, to the point D within it? BD and DC are less than the other two sides BA, AC of the triangle, but contain an angle BDC greater thao the angle BAC. Produce BD to E j and because two sides of a triangle are greater than tlje third side, the two sides BA, AE of the tri- 2 suigle OF EUCLID. 21 angle ABE are greater than BE. To each of these add EC j BookL therefore the sides BA, AC ^•^n'*^ are greater than BE, EC : A- gain, because the two sides CE, ED of the triangle CED are greater than CD, add DB to each of these ; therefore the sides CE, EB are greater than CD, DB ; but it has been shewn that B A, AC are greater than BE, EC, much more then are BA, AC greater than BD, DC. Again, because the exterior angle of a triangle is greater than the interior and opposite angle, the exterior angle BDC of the triangle CDE is greater than CED ; for the same rea- son, the exterior angle CEB of the triangle ABE is greater than BAC ; and it has been demonstrated that the angle BDC is greater than the angle CEB ; much more then is the angle BDC greater than the angle BAC. Therefore, if from the ends of, &c. Q E. D. PROP. XXII. PROB. X O make a triangle of which the sides shall besceN. equal to three given straight lines, but any two whatever of these must be greater than the third.* »20. i Let A, B, C be the three given straight lines, of which any two whatever are greater than the third, viz. A and B greater than C ; A and C greater than B j and B and C than A. It is required to make a triangle of which the side shall be equal to A, B, C, each to each. Take a straight line DE terminated at the point D, but un- limited towards E, and make * DF equal to A, FG to B, and GH equal to C; and from the cen- tre F, at the distance ^ FD, describe ''the circle'^ DKL ; and from the cen- tre G, at the distance GH, describe ^ another circle HLK j and join KF, KG; the triangle KFG has its sides equal to the three straight lines A, B, C. 3ecaus« the point F is the centre of the circle DKL, FD is C 3 «^ «> 3. Past. 21 THE ELEMENTS. Book t. equal ' to FK i but FD is equal to the straight line A ; there- ^"^S^viM fo''c FK *5 equal to A : Again, because G is the centre of the ' circle LKH, GH is equal " to GK -, but GPl is equal to C j therefore also GK is equal to C ; and FG is equal to B ; there- fore the three straight lines KF, FG, GK, are equal to the three A, B, C : And therefore the triangle KFG has its three sides KF, FG, GK. equal to the three given straight lines, A, B, C. Which was to be done. PROP. XXIII. PROB. jtVT a given point in a given straight line, to make a rectilineal angle equal to a given rectilineal angle. Let AB be the given straight line, and A the given point in it, and DCE the given rectilineal angle ; it is required to make an angle at the given point A in the q a given straight line AB, that shall be equal to the given rectilineal angle DCE. Take in CD, CE any points D, E, and "22. 1. join DE ; and make* the triangle AFG, the sides of which shall be eqiial to the three straight lines CD, DE, EC, so that CD be equal to AF, CE to AG, and DE to FG ; and because DC, CE are equal to F A, AG, each to each, and the base DE to the base FG j ^^- ■ • the angle DCE is equal ^ to the angle FAG. Therefore, at the given point A in the given straight line AB, the angle FAG is made equal to the given rectilineal angle DCE. Which was to be done. PROP. XXIV. THEOR. SeeN. 1 F two triangles have two sides of the one equal to two sides of the other, each to each, but the angle contained by the two sides of one of them greater than the angle contained b}^ the two sides equal to them, of the other; the base of that which has the greater angle shall be greater than the base of the . other. Let OF EUCLID. 2j Let ABC, DEF be two triangles which have the two sides ^0°"^ I- AB, AC equal to the two DE, DF, each to each, viz. AB ^^^^'^ equal to DE, and AC to DF ; but the angle BAC greater than the angle EDF ; the base BC is also greater than the base EF. Of the two sides DE, DF, let DE be the side which is not greater than the other, and at the point D, in the straight line DE, make * the angle EDG equal to the angle BAC j and * 23. l. make, DG equal ^ to AC or DF, and join EG, GF. ^ Because AB is equal to DE, and AC to DG, the two sides BA, AC are equal to the two ED, DG, each to each, and the angle BAC is equal to the angle EDG ; ^ j) therefore the baseN BC is "^ equal to thel base EG; and be-1 \ 1 W * ^' cause DG is equal to DF, the angle DFG is equal'' to the angle DGF ; but the angle DGF is greater than the an- gle EGF ; therefore the angle DFG is greater than EGF ; and much more is the an- gle EFG greater than the angle EGF j and because the angle EFG of the triangle EFG is greater than its angle EGF, and that the greater = side is opposite to the greater angle ; the side ' l?' EG is therefore greater than the side EF ; but EG is equal to BC ; and therefore also BC is greater than EF. Therefore, if two triangles, &c. Q^ E. D. PROP. XXV. THEOR. If two triangles have two sides of the one equal to two side^ of the other, each to each, but the base of the one greater than the base of the other ; the angle also contained by the sides of that which baa the greater base, shall be greater than the angle contained by the sides equal to them of the other. Let ABC, DEF be two triangles which have the two sides AB, AC equal to the two sides DE, DF, each to each, viz. AB equal to DE, and AC to DF ; but the base CB greater than the base EF j the angle BAC is likewise greater than the - angle EDF. C 4 For, 24 Book I, =■4. 1 '24. 1 THE ELEMENTS For, if it be not greater, it must either be equal to it,or less j but the angle BAC is not equal to the angle EDF, because then the base BC would beequal^toEF: jV but it is not ; there- fore the angle BAC is not equal to the angle EDF ; neither is it less ; because then the base BC would be less • ^ than the base EF ; but it Is not ; there- fore the angle BAC is not less than the angle EDF ; and it was shewn that it is not equal to it ; therefore the angle BAC is greater than the angle EDF. Wherefore, if two triangles, &c. Q^ E. D. PROP. XXVI. THEOR. J. F two triangles have two angles of one equal to two angles of the other, each to each ; and one side equal to one side, viz. either the sides adjacent to the equal angles, or the sides opposite to equal an- gles in each ; then shall the other sides he equal, each to each ; . and also the third angle of the one to the third angle of the other. Let ABC, DEF be two triangles which have the angles ABC, BCA equal to the angles DEF, EFD, viz. ABC to DEF, and BCA to EFD ; also one side equal to one side ; and first let those sides be equal which are adjacent to the an- gles that are equal in the two triangles ; viz. BC to EF ; the other sides . 1^ shall be equal, each ■ ■ to each, viz. AB to DE, and AC to DFj and the third angle BAC to lie third angle EDF. For, if A B be not B ^ C li V equal toDE, one of them must be the greater. Let AB be the greaterof thetwo, and make BG equal to DE, and joinGCj .therefore, because BG is equal to DE, and BC to EF, the two sides OF EUCLID. as sides GB, BC are equal to the two DE, EF, each to each ; and ^°°^ ^• the angle GBC is equal to the angle DEF ; therefore the base ^""'^'''^^ GC is equal* to the base DF, and the triangle GBC to the tri- a 4, 1. aqgle DEF, and the other angles to the other angles, each to each, to which the equal sides are opposite ; therefore the angle GCH is equal to the angle DFE; but DFE is, by the hypo- thesis, equal to the an;;le BCA j wherefore also the angle BCG is equal to the angle BCA, the less to the greater, which is im- possible: therefore AB is not unequal to DE,that is, it is equal to it ; and BC is equal to EF ; therefore the two AB, BC are equal to the two DE, EF, each to each j and the angle ABC is equal to the angle DEF ; the base therefore AC is equal* to the base DF, and the third angle B AC to the third angle EDF. Next, let the sides which are opposite to equal angles in each^ *^ triangle be equal to one another, viz. AB to DE ; likewise in this case, the other sides shall be equal, AC to DF, and BC to EF j and also the third angle BAC to the third EDF. For, if BC be not equal to EF, let BC be the greater of them, and make BH equal to EF, and join AH ; and because BH is equal to EF, and AB to DE j the two AB, BH are equal to the two DE, EF, each to each ; and they contain equal an- gles ; therefore the base AH is equal to the baseDF, and the triangle ABH to the triangle DEF, and the other angles shall be equal, each to each, to which the equal sides are opposite ; therefore the angle BHA is equal to the angle EFD ; but EFD is equal to the angle BCA ; therefore also the angle BHA is equal to the angle BCA, that is, the exterior angle BHA of the triangle AHC is equal to its interior and opposite angle BCA ; which is impossible'' ; wherefore BC is not unequal to" 16. 1. EF, that is, it is equal to it ; and AB is equal to DE ; therefore the two, AB, BC are equal to the two DE, EF, each to each ; and they contain equal angles j wherefore the base AC is equal to the base DF, and the third angle Bx^C to the third angle J^DF. Therefore, if two triangles, Sec. Q.E.D, •6 THE ELEMENTS Book t. PROP. XXVII. THEOR. If a straight line falling upon two other straight lines makes the alternate angles equal to one ano- ther, these two straight lines shall be parallel. Let the straight line E F, which falls upon the two straight lines AB, CD make the alternate angles AEF, EFD equal to one another; AB is parallel to CD. For, if it be not parallel, AB and CD being produced shall meet either towards B, D, or towards A, C : let them be pro- duced and meet towards B, D in the point G ; therefore GEF » 16. 1. is a triangle, and its exterior angle AEF is greater* than the interior and opposite angle EFG ; but it is also equal to it, which is impossible; there- fore AB and CD being pro- duced do not meet towards B, D. In like manner it may be demonstrated, that they do not meet towards A, C ; but those straight lines which meet neither way, though produced ever so far, are *55Pef. parajlel'' to one another. AB therefore is parallel to CD. Wherefore, if a straight line, &c. Q. E. D. PROP. XXVm. THEOR. jLF a straight line falling upon two other straight lines makes the exterior angle equ^l to the interior and opposite upon the same side of the line ; or makes the interior angles upon the same sidfc toge- ther equal to two Tight angles ; the two straight lines shall be parallel to one another. LetthestraightlineEFjWhich r, falls upon the two straight lines \^ AB,CD,maketheexterior angle \ EGB equal to the interior and « \(t opposite angle GHD upon the ' same side ; or make the interior angles on the same side BGH,^._ GHDtogether equal to two right angles ; AB is parallel to CD. Because the angle EGB is e- qual to the angle GHD, and lilie angle OF EUCLID. 27 angle EGB equal* to the angle AGH, the angle AGH is Boos i. equal to the angle GHD ; and they are the alternate angles ; i^^f"^ therefore AB is parallel^ to CD. Again, becaufe the angles "27. ]. BGH, GHD are equal<^ to right angles ; and that AGH, ' By Hip. BGH are alfo equal^ to two right angles ; the angles AGH," '^' ^• BGH are equal to the two angles BGH, GHD : Take away the common angle BGH ; therefore the remaining angle AGH is equal to the remaining angle GHD ; and they are alternate angles; therefore AB is parallel to CD. Wherefore if a ftraight line, &c. Q. E. D. PROP. XXIX. THEOR. If a straight line fall upon two parallel straight n^^^;^„ lines, it makes the alternate angles equal to one '!»'* p™p^ another; and the exterior angle equal to the inte- rior and opposite upon the same side ; and likewise the two interior angles upon the same side together equal to two right angles. Let the ftraight line EF fall upon the parallel ftraight lines AB, CD -J the alternate angles AGH, GHD are equal to one another ; and the exterior angle EGB is equal to the interior and oppofite, upon the fame fide, GHD J and the twointerior angles BGH, GHD upon the fame fide, are together equal to two right angles. For, if AGH be not equal to GHD, one of them muft be greater than the other ; let AGH be the greater; and becaufe the angle AGH is greater than the angle GHD, add to each of them the angle BGH ; therefore the angles AGH, BGH are greater than the angles BGH, GHD ; but the angles AGH, BGH are equal-* to two right angles ; therefore the » 13. 1. angles BGH, GHD are lefs than two right angles ; but those ftraight lines which, with another ftraight line falling upon them, make the interior angles on the fame fide lefe than two right angles, do meet* together if continually produced ; therefore * 12. Ax. the ftraight lines AB, CD, if produced far enough, ftiall meet ; ^^ ''"^ but they never meet, fmce they are parallel by the hvpothefis; thi) propo- thereforethe angle AGH is not unequal to the angle GHD, that '''''°''- is, it is equal to it ; but the angle AGH is equal"* to the angle b 15 j, EGB j therefore likewife EGB is equal to GHD ; add to each of THE ELEMENTS of thefe the angle BGH ; therefore the angles EGB, BGH are ('^^i]*^ equal to the angles BGH^ GHD ; but EGB, BGH are equal'^ to two right angles; therefore alfo BGH, GHD are equal to two right angles. Wherefore, if a ftraight, Sec. Q^ E. D. PROP. XXX. THEOR. oTRAIGHT lines which are parallel to the same straight line are parallel to one another. Let AB, CD be each of them parallel to EFj AB is alfo parallel to CD. Let the ftraight line GHK cut AB, EF, CD i and becaufe GHK cuts the parallel ftraight lines AB, EF, the angle AGH •J9. X. is equal* to the angle GHF. Again, becaufe the ftraight line GK. cuts the parallel ftraight lines EF, CD, the angle GHF is equal* jto the angle GKD ; and it was fliewn that the angle AGK is equal to the angle GHF ; therefore alfo AGK is equal to GKD j and they are *2?.l. alternate angles j therefore'AB is parallel'' to CD. . Where- fore ftraight lines, kc. Q^ E. D. PROP. XXXI. PROB. T. O draw a straight line through a given point pa- rallel to a given straight line. ■ Let A be the given point, and BC the given ftraight line •, it is required to draw a ftraight line through the point A, parallel to the {^^ ^ K ftraight line BC. '^ In BC take any point D, and join AD} and at the point A, in the f 23. 1. ftraight line AD, make* the angle tj ry DAE equal to the angle ADCj and produce the ftraight line EA to F. Becaufe the ftraight line AD, which meets the two ftraight lines BC, EF, makes the alternate angles EAD, ADC equal >^, 1. to one another, EF is parallel'^ to BC. Therefore the ftraight line OFEUCLID. 29 line E AF is 3rawn through the given point A parallel to the ^°°^ '• given ftraighc line BC. XVhich was to be done. ^ PROP. XXXII. THEOR. If a side of any triangle be produced, the exterior angle is equal to tlie two interior and opposite an- gles; and the tliree interior angles of every triangle are equal to two right angles. Let ABC be a triangle, and let one of its fides BC be produced co D ; the exterior angle ACD is equal to the two interior and oppofite angles CAB, ABC, and the three inte- rior angles of the triangle, viz. ABC, BCA, CAB, are toge- ther equal to two right angles, ThroughthepointCdraw CE parallel* to the ftrai^ht \ »3U 1. line AB ; and becaufe AB is parallel toCE,and AC meets them, the alternate angles B AC, ACE are equals A- y^ \ y^ bjg. 1. gain, becaufe AB is parallel ^ to CE, and BD falls upon '^ them,theexteriorangleECD is equal to the interior and oppofite angle ABC ; but the angle ACE was fhewn to be equal to the angle BAC ; therefore the whole exterior angl^ ACD is equal to the two interior and oppofite angles CAB, ABC ; to thefe equals add the angle ACB, and the angles ACD, ACB are equal to the three angles CB A, BAC, ACB ; but the angles ACD, ACB, are equal' to two right angles ; therefore c 13. i alfo the angles CBA, BAC, ACB are equal to two right an- gles. Wherefore if a fide of a triangle, &c. Q^ E. D. Cor. I. All the interior angles of any reflilineal figure, together with four right angles, are equal to ^ twice as many right angles as the ^ figure has fides. For any reftilineal figure ABCDE can be divided into as many triangles as the figure has fides, by drawing ftraight lines from a point F within the figure to each of its angles. And, by the preceding propofition, all 3» THE ELEMENTS Book J. a) J t^g angles of these triangles are equal to twice as many '"^'^'^'^''^ right apgles as there are triangles, that is, as there are sides of the figure ; and the same angles are equal to the angles of the figure, together with the angles at the point F, which is » 2 Cor. the common vertex of the triangles : that is", together with ^•'- ^- four right angles. Therefore all.the angles of the figure, to- gether with four right angles, are equal to twice as many right angles as the figure has sides. Cor. 2. All the exterior angles of any rectilineal figure are together equal to four rightangles. Because every interior angle ABC, with its adjacent exterior t-is. 1. ABD, is equal ^ to two right angles; therefore all the interior together with all the exterior angles of the figure, are equal to twice as inany tvj:\\t angles as there are sides of the figure ;^ that is, by the foregoing corol- lary, they are equal to all the interior angles of the figure, to- gether with four right angles •, therefore all the exterior angles are equal to four right angles. ^ * PROP. XXXIII. THEOR. . X HE straight lines which join the extremities of two equal and parallel straight lines, towards the same parts, arc also themselves equal and parallel. Let AB, CD be equal and pa- rallel straight lines, and joined^ towards the same parts by the straight lines AC, BD ; AC, BD are also equal and parallel. JoinBC; and because A B is parallel to CD, and BC meets C 1) ^- them, the alternate angles ABC, BCD are equal*; and because AB is equal to CD,and BC common to the two triangles ABC, DCB, the two sides AB, BC are equal to the two DC, CB ; and the angle ABC is equal to the angle BCD ; therefore the *• 1- base AC is cqual'^ to the base BD, and the triangle ABC to the triangle BCD, and the other angles to the other angles'", each to each, to which the equal sides are opposite ; therefore the angle » ng OF EUCLID. 3i angle ACB is equal to the angle CBD ; and becaufe the ^^ ftraight line BC meets the two itraight lines AC, BD, and makes the alternate angles ACB, CBD equal to one another, AC is parallel"^ to BD ; and it was ihewn to be equal to it. " - Therefore, ftraight lines, 6cc. Q_ E. D. PROP. XXXIV. THEOR. A HE opposite sides and anc^les of parallelograms are equal to one another, and the diameter bisects them, that is, divides them into two equal parts. N. B. A parallelogram is a four-sided Jigure, of u hich theoppositesidesarc parallel; and the diame- ter is the straight line joining tuo of its opposite angles. Let ABCD be a parallelogram, of which BC is a diame- ter ; the oppofice lides and angles of the figure are equal to one another ; and the diameter BC bifecls it. Becaufe AB is parallel to CD, a and BC meets them, the alter- nate angles ABC, BCD are equal *to one another; and becaufe » ^ \ aoo i AC is parallel to BD, and BC \ «y. i. meets them, the alternate angles ACB, CBD are equal* to one another ; wherefore the two triangles ABC, CBD have two angles ABC, BCA in one, equal to two angles BCp, CBD in the other, each to each, and one fide BC common to the two triangles, which is adjacent to their equal angled; there- fore their other fides {hall be equal, each to each, and the third angle of the one to the third angle of the other,'' viz. the fide'' AB to the fide CD, and AC to BD, and the angle BAC equal to the angle BDC : And becaufe the angle ABC is equal to the angle BCD, and the angle" CBD to the angle ACB, the whole angle ABD is equal to the whole angle ACD : And the angle BAC has been Ihewn to be equal to the angle BDC ; therefore the oppofite fides and angles of parallelograms are equal to one another ; alfo, their diameter bifecls thern ; for AB being equal to CD, and BC common, the two AB, BC are rqual to the two DC, CB, each to eachj and the aneL- ABC 32 THE ELEMENTS ^°^^ is equal to the angle BCD ; therefore the triangle ABC is '^4.i. equal'= to the triangle BCD, and the diameter BC divides the parallelogram ACDB into two equal parts. Q^ E. D. SeeN. Sec the 2d and 3d fi- gures, . »34, 1. 1 Ax. = 2. or 3, Ax. <29. 1. = 4. 1. • 3. Ax, PROP., XXXV. THEOR. -I Aral LELO GRAMS upon the same base, and be- tween the same parallels, are efjual to one another. Let the parallelograms ABCD, EBCF be upon the fame bafe BC, and between the fame parallels AF, BC ; the paral- lelogram ABCD fhali be equal to the parallelogram EBCF. it the fides AD, DF of the paral- lelograms ABCD, DBCF, oppofite a Ti TT to the bafe BC, be terminated in the ~ fame point D ; it is plain that each of the parallelograms is double* of the triangle BDC ; and they are there- fore equal to one another. - But, if the fides AD, EF, oppofite^ to the bafe BC of the parallelograms ABCD, EBCF, be not terminated in the fame point ; then, be- caufe ABCD is a parallelogram, AD is equal* to BC ; for the fame reafon EF is equal to BC j wherefore AD is equal'' to EF ; and DE is common ; therefore the whole, or the remain- der, AE is equal'^ to the whole, or the remainder DF ; AB alfo is equal to DC; and the two EA, AB are therefore equal to D F the two FD, DC, each to each ; and the exterior angle FDC is equal'' to the interior EAB, therefore the bafe EB is equal to the bafe FC, and the triangle E A B equal'^ to the triangle FDC : take the triangle FDC from the trapezium ABCF, and from the fame trapezium take the triangle EAB : the remainders therefore are equaF, that is, the parallelogram ABCD is equal to the parallelogram EBCF. Therefore parallelograms upon the fame bafe, &c. Q^ E. P. OF EUCLID. PROP. XXXVI. THEOR. X ARALLELOGRAMS upon equal bases, and between the same parallels, are equal to one another. Let ABCD, EFGH be ^ p. ^ parallelograms upon equal bases BC, FG, and be- tween the same parallels AH, BG ; the paraHelo- gram ABCD is equal to EFGH. i^ ,, ^ J.vin BE, CH ; and be- '> ^ ^ G cause BC is equal to FG, and FG to' EH, BC is equal to *54. i. EH : and thev are parallels, and joined towards the same parts by t'le straight lines BE, CH : But straight lines which join equal and parallel straight lines towards the same parts, are themselves equal and parallel ;'' therefore EB, CH, are both "33.1. equal and parallel, and EBCH is a parallelogram j and it is equal*^ to ABCD, because it is upon the same base BC, and "^5. i, between the same parallels ^C, AD : For the like reason, the parallelogram EFGH is equal to the same EBCH : There- fore also the parallelogram ABCD is equal to EFGH. — Wherefore parallelograms, &c. Q. E. D. PROP. XXXVn. THEOR. I RI ANGLES upon the same base, and between the same parallels, are equal to one another. Let the triangles ABC, DBC, be upon the same base BC,and between the same parallels 17 \ 1> V AD, BC: The triangle ABC -' -^^^ — — ^ is equal to the triangle DBC. Produce AD both ways to the points E, F, and through \ / / \ \ / ^ . B draw^ BE parallel to CA ; ^ ^^ ^^ ^ *^^' ' and through C draw CF pa- rallel to BD : Therefore each of the figures EBCA, DBCF is a parallelogram ; and EBC A is equal^ to DBCF, because they are upon the same .base BC, " 35. 1. and. between the same parallels BC, EF ; and the triangle ABC is the half of the parallelogram, EBCA, because the D diameter THE ELEMENTS diameter AB bisects <= it ; and the triangle DBC is the half of the parallelogram DBCF, because the diameter DC bisects it : but the halves of equal things are equal i*^ therefore the triangle ABC is equal to the triangle DBC. Wherefore triangles, Sec. Q^ E. D. PROP. XXXVIII. THEOR. J. RIANGLES upon equal bases, and between the same parallels, are equal to one another. Let the triangles ABC, DEF be upon equal bases BC, EF, and between the same parallels BF, AD : The triangle ABC Js equal to the triangle DEF. Produce AD both ways to the points G, H, and through B draw BG parallel * to CA, and through F draw FH paral- lel to ED : Then each p a y>. Vi of the figures GBCA,f ~^^ -^ ^^ DEFH, is a parallel- ogram; and they are * 36. 1. equal to'= one another, because they are upon equal bases BC, EF, and between the same t5 i.- IL r < 34. 1. parallels BF, GH ; and the triangle ABC is the half *^ of the parallelogram GBCA, because the diameter AB bisedts it ; and the triangle DEF is the half "^ of the parallelogram DEFH, because the diameter DF bise£ls it : But the halves of equal •^ 7. Ax. things are equal ;'' therefore the triangle ABC is equal to the triangle DEF. Wherefore triangles, &c. Q^ E. D. •31. 1. PROP. XXXIX. THEOR. Equ. lLQUAL triangles upon the same base, and upon the same side of it, are between the same parallels. Let the equal triangles ABC, DBC be upon the same base BC, and upon the same side of it ; they are between the same parallels. Join AD ; AD is parallel to BC ; for, if it is not, through • 31.-1. the point A draw^ AE parallel to BC, and join EC : The X triangle I OF EUCLID. triangle ABC is equal '' to the triangle a EBC, because it is upon the same base ^ BC, and between the same parallels EC, AE : But the triangle ABC is equal to the triangle BDC ; therefore also the triangle BDC, is equal to the triangle EBC, the greater to the less, which is impossible : Therefore AE is '^ ^ not parallel to BC. In the same manner, it can he demon- strated, that no other line but AD is parallel to BC ; AD is therefore parallel to it. Wherefore equal triangles upon, &c. Q. E. D. 35 Book I. PROP. XL. THEOR. E. QIJAL triangles upon equal bases, in the same straight line, and towards the same parts, are be- tween the same parallels. Let the equal triangles ABC, DEF be upon equal bases BC, EF, in the same straight ^ D line BF, and towards the same parts ; they are be- tween the same parallels. Join AD ; AD is paral- lel to BC : For, if it is not, through A draw^ AG parallel to BF, and join GF : The triangle ABC is equal'' to the triangle GEF, be- i cause they are upon equal bases BC, EF, and between the same f>arallels BF, AG : But the triangle ABC is equal to the triangle DEF ; therefore also the triangle DEF is equal to the triangle GEF, the greater to the less, which is impossi- ble : Therefore AG is not parallel to BF : And in the same manner it can be demonstrated that there is no other parallel to it but AD, AD is therefore parallel to BF. Wherefore equal triangles, kc. Q. E. Dt »31. 1. 33. 1. PROP. XLI. THEOR. . F a parallelogram and triangle be upon the same base, and between the same parallels ; the parallel- ogram shall be double of the triangle. D2 Let 36 Book I. •37. 1. "34.1, THE ELEMENTS Let the parallelogram ABCD and the triangle EBC be upon the same base BC, and between the same parallels BC, AE ; the parallelogram ABCD is double of ^ T\ -p the triangle EEC. "^ ' Join AC ; then the triangle ABC is equal ^ to the triangle EBC, because they are upon the same base BC, and between the same parallels BC, AE. But the parallelogram ABCD is double'' of the triangle ABC, because the diameter AC divides it into two ; " C equal parts ; wherefore ABCD is also double of the triangle EBC. > Therefore, if a parallelogram, &c. Q_. E. D. PROP. XLIL PROB. » 10. 1. •^31.1. "38. 1. «4i. 1. JL O describe a parallelogram that shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle. Let ABC be the given triangle, and D the given reftilineal angle. It is required to describe a parallelogram that shall be equal to the given triangle ABC, and have one of its angles equal to D. Bisect^ BC in E, join AE, and at the point E in the straight line EC make'' the angle CEF equal to D ; and through A draw*^ AG parallel to EC, and * v r* through C draw CG"= parallel to '^~* ' EF : Therefore FECG is a parallelogram: And because BE is equiil to EC, the triangle ABE is likewise equal'' to the triangle AEC, since they are upon equal bases BE, EC, and between the /^ same parallels BC, AG ; there- H fore the triangle ABC is double of the triangle AEC. And the parallelogram FECG is likewise double*' of the triangle AEC, because it is upon the same base, and between the san>e parallels : Therefore the parallelogram FECG is equal to the triangle ABC, and it has one of its angles CEF equal to the give-a angle D j wherefore there has been described a parallel^. ogram OF EUCLID. 37 ©gram FECG equal to a given triangle ABC, having one of ^<">^ I- its angles CEF equal to the given angle D. Which was to ^"^"^^^^ be done. PROP. XLIII. THEOR. A HE complements of the parallelograms, which are about the diameter of any parallelogram, ^re equal to one another. LwC ABCD be a parallelogram, of which the diameter is AC, and LH, FG, and parai- ^.^ J^ lelograms ao lut AC, that is through ivhLh AC passes.^ and BK, KD, the other parallelo- grams wh ch make up the whole figure ABCD, which are there- fore called the complements : The complement BK is equal ^ to the complement KD. Vi G ' C Because ABCD is a parallelogram, and AC ks diameter, , the triangle ABC is equal' to the triangle ADC : And, because * " ^' EKHA IS a parallelogiam, the diametec of which is AK, the triangle AEK is equal to the triangle AHK : By the same reason, the triangle KGC is equal to me triangle KFC: Then, because the trianglr AEK is equal to the triangle AHK, and the triangle KGC to KFC ; the triangle AEK, together with the triangle KGC is equal to the triangle.AHK together with the triangle KFC : But, the whole triangle ABC is equal to the whole ADC i ihereiore the remaining complement BK is equal to the remaining complement KD. Wherefore the complements, &c. Q^ E. D. PROP. XLIV. PROB. X O a given stmight line to apply a parallelogram, which shall be equal to a given tnangle, and have one of its angles equal to a given rectilineal angle. Let AB be the given straight line, and C the given triscn- gle, and D the given rectilineal angle. It is required toap- ply to the straight line AB a parallelogram equal to the tri- angle C, and having an angle equal to D. D3 Make 38 THE ELEMENTS »>31. 1. 29. 1. « 12. Ax ^43. 1. n5. 1. , Make* the parallelogram BEFG equal to the triangle C, and hav- ing the angle EBG equal to the angle D, so that BE 'he in the same straight line with AB, and produce FG to H ; and through A draw'' AH parallel to BG or EF, and join HB. T^hen be- ' cause the straight line HF falls upon the parallels AH, EF, the angles AHF, HFE, are together equa^ to two right an- gles ; wherefore the angles BHF, HFE are less than two right angles : But straight lines which with another straight line tnake the interior angles upon the same side less than two right angles, do meet'' if produced far enough : Therefore HB, FE shall meet if produced ; let them meet in K, and through K draw KL, parallel to EA or FH, and produce HA, GB to the points L, M : Then HLKF is a parallelogram, of which the diameter is HK, and AG, ME are the parallelograms about HK ; and LB, BF are the complements: therefore LB is equal^ to BF ; But BF is equal to the triangle C ; wherefore LB is equal to the triangle C ; and because the angle GBE is equal*^ to the angle ABM, and likewise to the angle D ; the angle ABM is equal to the angle D : Therefore the parallelogram LB is applied to the straight line AB, is equal to the triangle C, and has the angle ABM equal to the angle D : Which, was to be. done. SeeN, »42. 1. '44.1. PROP. XLV. PROB. L O describe a parallelogram equal to a given rec- tilineal figure, and having an angle equal to a given rectilineal angle. Let ABCD be the given redilineal figure, and E the given rectilineal anpjle. It is required to describe a parallelogram equal to ABCD, and having an angle equal to E. Join DB, and describe'' the parallelogram FH equal to the triangle ADB, and having the angle HKF equal to the angle E i and to' the straight line GH apply^ the parallelogram GM equai OF EUCLID. 39 2i>. 1. equal to the triangle DBC, having the angle GHM equal to BookI. the angle E ; and because the angle E is equal to each of the ^'^^■''^^ angles FKH, GHM, the angle EKH is equal to GHM : add to each of these the angle KHG ; therefore the angles FKH, KHG are equal A 1) p^ G L, to the angles KHG, GHM; butFKH,KHG are equal '^ to two right angles; Therefore also KHG, GHM, are equal to two right angles; and because at the point H in the straight line GH, the two straight lines KH, HM, upon the opposite sides of it make the adjacent angles equal to two right angles, KH is in the same straight line** with HM ; and because the straight line " U. i. HG meets the parallels KM, FG, the alternate angles MHG, HGF are equal := Add to each of these the angle HGL : Therefore ihe angles MHG, HGL, are equal to the angles HGF, HGL : But the angles MHG, HGL, are equal " to two right angles ; wherefore also the angles HGF, HGL are equal to two right angles, and FG is therefore in the same straight line with GL ; and because KF is parallel to HG, and HG to ML ; KF is parallel to ML ; and KM, FL are pa-^so i rallek ; wherefore KFLM is a parallelogram ; and because the triangle ABD is equal to the parallelogram HF, and the tri- angle DBC to the parallelogram GM; the whole reililineal figure ABCD is equal to the whole parallelogram KFLM ; therefore the parallelogram KFLM has been described equal to the given reftilineal figure ABCD, having the angle FKM equal to the given angle E. Which was to be done. Cgr. From this it is manifest how to a given straight line to apply a parallelogram, which shall have an angle equal to a given re£Vilineal angle, and shall be equal to a given rectilineal figure, viz. by applying'' to the given straight line a parallel- 1> 4^. :. ogram equal to the first triangle ABD, and having an angle «qual to the given angle. 04 40 THE ELEMEI^TS Book I. PROP. XLVI. PROB. X O describe a square upon a given straight line. Let AB be the given straight line; It is required to de- scribe a square upon AB. Ml. 1. Frdm the point A draw^ AC at right angles to AB ; and '3. ], make^ AD equal to AB, and through the point D draw DE «3i. 1, parallel to AB, and through B draw BE parallel to AD ; <i34 1 therefore ADEB is a parallelogram : whence AB is equal'' to DE, and AD to BE : But B A is equal (^ to AD; therefore the four straight lines BA, AD, DE, EB are equal to one -another, and the parallelogram ADEB -t) is equilateral, likewise all its angles are right angles ; because the straight line AD meeting the parallels AB, DE, ^29.1. the angles BAD, ADE are equal-^ to two right angles ; but BAD is a right angle ; therefore also ADE is a right angle ; but the opposite angles of paral- Al B lelograms are equal ;•* therefore each of the opposite angles ABE, BED is a -right angle*; wherefore the figure ADEB is rectangular, and It has been demonstrated that it is equila- teral; it is therefore a square, and it is described upon- the given straight line AB ; Which was to be done. Cor. Hence every parallelogram that has one right angle has all its angles right angles. PROP. XLVII. THEOR. J. NT any right-angled triangle, the square which is described upon the side subtending the right angle, is equal to the squares described upon the sides which contain the right angle; Let ABC be a right-angled triangle having 'the right angle BAC ; the square described upon the side BC is equal to the squares described upon BA, AC. ''^- 1- On BC describe'' the square BDEC, and on BA, AC the square? «fu;». OF EUCLID. squares GB, HC ; and through A draw *> AL parallel to BD, or CE, and join AD, FC ; then, because each of the angles BAG, BAG is a right angles the two straight lines AC, AG, upon the opposite side* of A B, make witn ii ac the point A the adjacent angles ' equal to two right angles ; therefore CA is in the sarrie straight line ^ with AG; for the same reason, AB and AH are in the same straight line ; and because the angle DBC is equal to the ar^gle FBA, each ot them being a right annle, add to each the angle ABC, and the whole angle DBA is equal ^ to the whole FBC ; and ' 2 Ak. because the two sides AB, BD are equal to the two FB, BC, each to each, and the angle DBA equal to the angle FBCj therefore the base AD is equal ^ to the base FC, and the trian- f 4. i. gle ABD to the triangle FBC : Now the parallelogram BL is double E of the triangL ABD, because they are upon the saoie s 41. i. base BD, and between the same parallels, BD, AL ; and the square G3 is double of the triangle FBC, because these also are upon the same base FB, and between the same parallels FB, GC. But the doubles of equals are equal •" to one ano- » 6. .\jt ther ; Therefore the parallelogram BL is equai to the square GB: And, in the same manner, by J9ining AE, BK, it is de- monstrated, that the parallelogram CL is equal to the square HC ! Therefore the whole square BDEC is equal to the two squares GB, HC ; and the square BDEC is described upon the straight line BC, and the squares GB, HC upon BA, AC : Wherefore the square upon the side BC is equal to the squares upon the sides BA, AC Therefore, in any right-angled tri- angle, &c. Q, E. D. PROP. XLVIIL THEOR. If the square described upon one of the sides of a triangle, be equal to the squares describedupon the other two sides of it ; the aujrle contained bv these two sides is a riiiht an«:le. If 42 Book I. • JI. 1. 47.1. ^6. 1. THE ELEMENTS, &c. If the square described upon BC, one of the sides of the tri- angle ABC, be equal to the squares upon the other sides BA, AC, the angle BAC is a right angle. From the point A draw* AD at right angles to AC, and make AD equal to BA, and join DC : Then, because DA is equal to AB, the square of DA is equal to the squares of AB : To each of these add the square of AC ; therefore the squares of Da, AC are equal to the squares of BA, AC : But the square of DC is cqual^ to the squares of DA, AC, because DAC is a right angle ; and the square of BC, by hypothesis, is equal to the squares of BA, Ac ; therefore the square of DC is ^^ equal to the square of BC ; and therefore o also the side DC is equal to the side BC. And because the side DA is equal to AB, and AC common to the two triangles DAC, BAC, the two DA, AC are equal to the two BA, AC j and the base DC is equal to the base BC ; therefore the angle DAC is equal <= to the angle BAC ; but DAC is a right an-^ gle; therefore also BAC is a right angle. Therefore, if the S(juare, &c. Q. E. D, THE Book. 11. ELEMENTS EUCLID, BOOK. II. DEFINITIONS. I. ijjVERY right angled parallelogram is said to be contained by any two of the straight lines which contain one of the right angles. II. In every parallelogram, any of the parallelogranis about a diju. meter, together with the two complements, is called a Gno- mon. ' Thus the parallelo- ' gram HG, together with the * complements AF, FC,is the * gnomon, which is more * briefly expressed bvthe let- < ters AGK, or EHC, which * are at the opposite angles of ^ * the parallelograms which make the gnomon.' PROP. I. THEOR. If there be two straight lines, one of which is di- vided into any number of parts ; the rectangle con- tained by the two straight lines, is equal to the rec- tangles contained by the undivided line, and the several parts of the divided line. n. 1. *3. J. '31. 1. 54. ]. THE ELEMENTS Let A and BC be two straight lines ; and let BC be divided into any partsin the points D, E j the rectangle contained by B the straight lines A, BC is equal to the rectangle contained by A, BD, together with that con- tained by A, DE, and that con- tained by A, EC. From the point B draw^ BF ; Q at right angles to BC,ancl make BG equal '" to A ; and through G draw "^ GH parallel to BC ; and through D, E, C, draw ^ DK, EL, CH parallel to BG; then the rectangle BH is equal to the rectangles BK, DL, EH ; and BH is contained by A, BC, for it is contained by GB, bC, and GB is equal to A ; and BK is contained by A, BD, for it is contained by GB, BD, of which GB is equal to A ; and DL is contained by A, DE, because DK, that is ^ BG, is equal to A ; and in like manner the rect- angle EH is contained by A, EC : Therefore the rectangle contained by A, BC, is equal to the several rectangles con- tained by A, BD, and by A, DE ; aiid also by A, EC. Wherefore, if there be two straight lines, Sic. Q^ E, D. •4i>. 1, . PROP. IL THEOR. XT a straight line be divided into any two parts, the rectangles contained by the whole and each of the parts, are together equal to the square of the whole line. Let the straight line AB be divided into any two parts in the point C; the rectangle contained by AB, BC ; toge- ther with the rectangle* AB, AC, shall be equal to the square of AB. Upon AB ^describe » the sqbare ADEB, and' through C draw ^ CF, parallel to AD or BE ; then AE is equal to the rectangles AF, CE ; and AE is _ ^ ^ the square of AB j and AF is the rectangle contained'by BA, AC ; for it is contained by DA, AC of which AD is equal B. To *N . B. To avoid repratini; the word contained Xoo freqoentlv, tl.c rectangle ' by two straight lines AB. AC is sometimes simply called t!ie rectangle AB, con- AC. to OF EUCLID. 45 to AB J and CE is contained by AB, BC, for BE is equal to ^^^^ AB ; therefore the rectangle contained by AB, AQ together ''^"'''''^ with the rectangle AB, BC, is equal to the square of AB. If therefore a straight line, ice. Q: E. D. PROP. III. THEOR. F a straight line be divided into any two parts, the rectangle contained by the whole and one of the parts, is equal to the rec tingle contained by the two parts, together with the square of the aforesaid part. Let the straight line AB be divided into two parts in the point C ; the rectangle AB, BC, is equal to the recungle AC, CB, together with the square of BC. Upon BC describe * the square A C B * ^^- ^ CD£B, and produce ED to F, and through A draw^'AF parallel to i>5i_ j^ CD or BE ; then the rectangle AE is equal to the rectangles AD, CE i ajwJ AE is the rectangle contained by AB, BC, for it is contained by, AB, BE, of which BE is equal to BC ; and AD is contained by AC, V CB, for CD is equal to BC ; and DB is the square of BC ; therefore the rectangle AB, BC, is equal to the rectangle AC, CB, together with the square of BC. If therefore a straight line, &c. Q. E. D. PROP. IV. THEOR. 1 F a straight line be divided into any two parts, the square of the whole line is equal to the squares of tlie two parts, together with twice the rectangle contained by the parts. Let the straight line AB be divided into any two parts in C ; the square of AB is equal to the squares of AC, CB, and to twice the rectangle contained by AC, CB. Upon ■/ ^ 46 Book II. »46. 1. «>31. 1. ' 29. 1. «»5. 1. e6. 1. <"34.. 1. e43. 1. THE ELEMENTS Upon AB describe =» the square ADEB, and join BD, ancJ through C draw*- CGF parallel to AD or BE, and through G draw HK parallel to AB or DE: And because CF is parallel to AD, and BD falls upon them, the exterior angle BGC is equal"^ to the interior and opposite angle ADB ; but ADB is equal <^ to the angle ABD, because BA is equal to AD, being sides of a square ; wherefore the an- gle CGB is equal to the angle GBC ; and therefore the side BC is equal ^ to the side CG: But CB is equal* also to GK, and CG to BK ; wherefore the figure CGKB is equilateral : It is likewise rectangular ; for CG is pa- rallel to BK, and CB meets them j the angles KBC, GCB are therefore equal to two right angles ; and KBC is a right angle ; where- fore GCB is a right angle ; and therefore also the angles ^ CGK, GKB opposite to these, are right angles, and CGKB is rectangular; but it is also equilateral, as was demonstrated ; wherefore it is a square, and it is upon the side CB : For the same reason HF also is a square, and it is upon the side HG, which is equal to AC : Therefore HF, CK are the squares of AC, CB ; and because the complement AG is equal e to the complement GE, and that AG is the rectangle con- tained by AC, CB,. for GC is equal to CB ; therefore GE is also equal to the irectangle AC, CB ; wherefore AG, GE are equal to twice the rectangle AC, CB : And HF, CK are the squares of AC, CB ; wherefore the four figures HF, CK, AG, GE are equal to the squares of AC, CB, and to twice the rectangle AC, CB : But HF, CK, AG, GE make up the whole figure ADEB, which is the square of AB : There- fore the square of AB is ^qual to the squares of AC, CB, and twice the rectangle AC, CB. Wherefore if a straight line, &c. Q, E. D. CoR. From the demonstration, it is manifest, that the pa- rallelograms about the diameter of a square arc likewise squares. OF EUCLID. 47 Book II. PROP. V. THEOR. JtF a straight line be divided into tw^o equal parts, and also into two unequal parts, the rectangle con- tained by the unequal parts, together with the square of the line betAveen the points of section, is -equal to the square of half the line. Let the straight line AB be divided into two equal parts in the point C, and into two unequal parts at the point D; the redangle AD, DB, together with the square of CD, is equal to the square of CB. Upon CB describe" the square CEFB, join BE, and through * -i^- ^• Ddraw" DHG parallel to CE or BF ; and through H draV ■> 31. i. KLM parallel to CB or EF ; and also through A draw AK parallel to CL or BM : And because the complement CH is equal'-" to the complement HF, to each of these add DM; there- c 43. 1. fore the whole CM is equal J^ to the whole DF ; but CM '" Is equaH to AL, because AC is equal to CB ; there- }^ fore also AL is equal to DF. To each of these add CH, and the whole AH is equal to DF and CH : But AH E G ¥ is the redangle contained by AD, DB, for DH is equal = tO'Cor. 4. 2. DB i and DF together with CH is the gnomon CMG ;' therefore the gnomon CMG is equal to the rectangle AD, DB : To each of these add LG, which is equal= to the square of CD ; therefore the gnomon CMG, together with LG, is equal to the re6tangle AD, DB, together with the square of CD : But the gnomon CMG and LG make up the whole figure CEFB, wiiich is the square of CB : Therefore the retlangle AD, DB, together with the square of CD, is equal to the square of CB. Wherefore^' if a straight line, &c. Q, E. D. From this proposition it is manifest, that the difference of the squares of two unequal lines AC, CD, is equal to the re<5langle contarned by their sum and difference. 36. 1. ^ Book U. TriE ELEMENTS PROP. VI. THEOR. •46. 1. •31. L «45. 1 J.F a straight line be bisected, and produced to any point; the rectangle contained by the whole line thus produced, and the part of it produced, toge- ther with tiie square of half the line bisected, is equal to the square of the straight line which is made up of the half and tlie part producetl. Let the straight line AB be bisected in C, and produced to the point D ; the reSangle AD, DB, together with the square of CB^ is equal to the 'square ot CD. UponCDdescribeHhe square CEFD,join DE,and through B draw'' BHG parallelto, CE or DF, and through H draw KLM parallel to AD or EF, and also through A draw AK pa- rallel to CL or DM; and be- ^ (3 ^ J) cause AC is equal to CB, the re6langle AL is equal^ to CH ; but CH is equaF toy^ HF : therefore also AL is equal to HF: To each of these add CM; therefore the whole AM is equal to the gnomon CMG: And AM is the < Cor. 4.2. reilangle contained by AD, DB, for DM is equa^ to DB : Therefore the gnomon, CMG is equal to the rc6langle AD, DB : Add to each of these I/G, which is equal to the square of CB, therefore the reftangle AD, DB, together with the^ square of C3, is equal to the gnomon CMG, and the figure LG; But the gnomon CMG and LG make up the whole figure CEFD, which is the square of CD ; therefore the redlangie AD, DB, together with the square of CB, is equal to the square oi CD. W herefore, if a straight line, Sec. Q^ E. D. PROP. VIL THEOR. If a straight line he divided into any two parts, the squares of the whole line, and -of one of the parts, arc equal to twice the rectangle contained by the whole and that part, together with the square of the other part. Let the straight line AB be divided into any two parts in the OF EUCLID. 49 the point C; the squares of AB, EC are equal to twice the Book.II. reftangle AB, BC, together with the square of AC. ^.-"v^ ' Upon AB describe * the square ADtB, and construft t^e* ^o. i. figure as in the preceding propositions; and because AG is equal'' to GE, add to each of them CK ; the whole AK is* 40. i. therefore, equal to the whole CE j therefore AK, CE, are double of AK : But AK, CE, are the gnomon A AKF, together with. the square CK ; therefore the gnomon AKF, toge- ^ ther with the square CK, is double of AK : But twice the rectangle AB, BC is double of AK, for BK is equa^ to BC : Therefore the gno- mon AKF, together with the square j^"^ CK, is equal to twice the re^angle AB, EC : To each of these equals add HF, which is equal to the square of AC ; therefore the gnomon AKF, together with the squares CK, HF, is equal to twice the re<5langle AB, BC, and the square of AC : but the gnomon AKF, together with the squares CK, HF, make up tiie whole figure ADEB and CK, which are the squares of AB and BC : therefore the squares of AB and BC are equal to twice the rectangle AB, BC, together with the square of AC. Wherefore, if a straight line, &c. Q. E. D. « Cor. 4. C. PROP. VIII. THEOR. J.F a straight line be divided into any two parts, four times the rectangle contained by the whole line, and one of the parts, together with the square of the other parr, is equal to the square of the straight line, which is made up of the whole and that part. Let the straight line AB be divided into any two parts in the point C-, four times the redangle AB, BC, together with the square of AC, is equal to the square of the straight line made up of AB and BC together. Produce AB to D, so that BD be equal to CB, and upon AD describe the square AEFD ; and construct two figures such as in the preceding. Because CB is equal to BD, and that CB is equal* to GK, and BD to KN ; therefore GK is »34. £ equal 50 Book IT. THE ELEMENTS •43.1. X equal to KN : For the same reason, PR is equal to RO ; and because CB is equal to BD, and GK to KN, the redlangle CK is equal'' to BN, and GR to RN : but CK is equal^ to RN, because they are the complements of the parallelogram CO ; therefore also BN is equal to GR ; and the four redit- angles BN, CK, GR, RN are therefore equal to one another, and so are quadruple of ont of them CK : Again, because CB is equal to BD, and that BD is ^ Cor. 4. 2. equaH to BK, that is, to CG , and CB equal to GK, that ^ is, to a GP ; therefore CG is equal to GP : And because CG is equal to T\/r ' GP, and PR to RO, the reftangla ^ ^ AG is equal to MP, and PL to RF: But MP is equal ^ t» PL, because they are the complements of the parallelogram M L ; where- fore AG is equal also to RF : Therefore the four redlangles AG, MP, PL, RF are equal to one another, and so are quadruple of one of them AG. And it was demonstrated that the four CK BN, GR, and RN are quadruple of CK. Therefore the eight rectangles which contain the gnomon AOH, are quad- ruple of AK ; .and because AK is the rectangle contained by AB, BC, for BK is equal to BC, four times the redangle AB, BC is quadruple of AK : But the gnomon AOH was demonstrated to be quadruple of AK : therefore four times the rectangle AB, BC, is equal to the gnomon AOH. To each of these add XH, which is equaK to the square of AC : Therefore four times the rectangle AB, BC together with the squareof AC, is equal to the gnomon AOH and the square XH: But the gnomon AOH and XH make up the figure AEFD, which is the square of AD : Therefore four times the redtangle AB, BC, together with the square of AC, is equal to the square of AD, that is, of AB and BC added together in one straight line. Wherefore, if a straight line, &c. Q;. E. D. ^ Cor. 4.2. OF EUCLID. 5, Book II. . PROP. IX. THEOR. J.F a straight line be divided into two ecjual, and also into two unequal parts; the squares of the tv/o unequal parts are together double of the square of half the line, and of the square of the line between the points of section. Let the straight line Ah be divided at the point C into two equal, and at D into two unequal parts : The squares of AD, DB are together double of the squares of AC, CD. From the point C draw* CE at right angles to AB, and Mi. ji. make it equal to AC or CB,and join E A, EB ; through D draw "DF parallel to CE, and through F draw FG parallel to AB ; " si. }. and join x\F ; Then, because AC is equal to CE, the angle EAC is equal<= to the angle A EC ; and because the angle "^ 5.1. ACE is a right angle, the two others AEC, EAC together make one right arigle"* ; and they are equal to one another •,"22.1. each of them therefore is half of a right angle. For the same reason each of the angles CEB, EBC is half a right angle j and therefore the whole AEB is a right angle : And because the an- gle GEF is half a right angle, and EGF a right angle, for it is cquaP to the interior and opposite angle ECB, the remaining eoo. 1, angle EFG is half a right angle ; therefore the angle GEF is equal to the angle EFG, and the side EG equaF to thcfg, j, side GF; Again, because the angle at B is half a right angle, and FDB a right angle, for it is equal ' to the in- terior and opposite angle ECB, the remaining angle BFD is half a right angle ; therefore the angle at B is equal to the angle BFD, and the side DF to ^ the side DB : And because AC is equal to CE, the square of AC is equal to the square of CE i therefore the squares of AC, CE, are double of the square of AC : But the square of EA is equals to the squares evT. 1. of AC, CE, because ACE is a right angle; therefore the square of EA is double of the square of AC : Again, be- cause EG is equal to GF, the square of EG is equal to the square of GF ; therefore the squares of EG, GF are double of E 2 the J \ A ( > D B 47. 1. 52 THE ELEMENTS Book u. the squarc of GF i but the squire of EF is equal to the squares ^"^'"''''^^ of EG, GF ; therefore the square of EF is double of the square * 34. 1. Qp . 2j^(j Qp js Qqualh to CD ; therefore the square of EF is double of the squarc of CD : But thesguare of AE is likewise double of the square of AC ; therefore the squares of AE, EF are double of the squares of AC, CD : And the square of AF is equal' to the squares of AE, EF, because AEF is a right angle j therefore the square of AF is double of the squares of AC, CD : But the squares of AD, DF, are equal to the square of AF, because the angle ADF is a right angle j therefore the squares of AD, DF are double of the squares of AC, CD : And DK is equal toDB; therefore the squares of AD, DB arc double of the squares of AC, CD. If therefore a straight line. Sec. Q.E. D. PROP. X. THEOR. IF a straight line be bisected, and produced to any point, the square of the whole line thus pro- duced, and the square of the part of it produced, are together double of the square of half the line bisected, and of the square of the hne made up of the half and the part produced. Let the straight line AB be biseded in C and produced to the point D ; the squares of AD, DB are double of the squares ■ of AC, CD. * n. 1. From the point C draw* CE at right angles to A, B : And make it equal to AC or CB, and join AE, EB ; through E draw *3l. 1. bgp parallel to AB, and through D draw DF parallel to CE: And because thestraight line EFmeets the parallelsEC,FD, the . 29 J angles CEF, EFD are equa^to two right anglts; and therefore theanglesBEF,EFDarelessthantwor!ghtangles; butstraight lines which with another straight line make the interior angles 12. Ax. upon the same side less than two right angles, do meet"* if pro- duced far enough : Therefore EB, FD shall meet, if produced ^ towards B,D: Let them meet in G, and join AG: Then, because AC is equal to CE, the angle CEA is equal* to the angle EAC ; and the angle ACE is a right angle ; therefore each of '3i- i- the angles CEA, EAC is half a right angle'. For the same reason, OF EUCLID, 53 «6. 1, t34. 1, reason, each of the angles CEB, EEC is half a right angle; ^°°^ therefore AEB is a right angle : And because EBC is half a ^^"^'^^ right angle, DBG is also^ half a right angle, for they are^ ver- '^^s, i. tically qppositc i but BDG is a right angle, because u is equalise. 1. to the alternate angle DCE j therefore the remaining angle DGB is half a right angle, and is therefore equal to thd angle DBGj wherefore also the side BD is equal? to the side DG Again, because EGF is half a right angle, and that fc^ £ . the aagle at F is a right an- gle, because ii is equal"* to the oD-)oS4teangleECD, jthc -emaining angle FEG Ss :ulf a V.ght angle, and \4 ej ial to tie angle EGF ; v- efsfore also the side CF is equals to the side F£. And because EC is equal to CA, the square of EC is equal to the square of CA ; therefore the squares of EC, CA are Jouble of the square of CA : But the square of EA is equal' to the squares of EC, C A ; therefore the square of EA * ^^ is double of the square of AC : Again, because GF is equal to FE, the square of GF is equal R) the square of FE ; and therefore the squares of GF, FE are double of the square of EF ; But the square of EG is equal- to the squares of GF, FE; therefore the square of EG is double of the square of EF : And EF is equal to CD ; wherefore the square of EG is dou- ble of the square of CD . But it was demonstrated, that the square of EA is double of the square of AC ; therefore the squares of AE, EG, are double of the squares of AC, CD ; And the square of AG is equaU to the squares of AE, EG ; therefore the square of AG is double of the squares of AC, CD : But the squares of AD, GD, are equal' to the square of AG i therefore the squares of AD, DG are double of the squares of AC, CD : But DG is equal to DB ; therefore the squares of AD, DB are double of the squares of AC, CD. Wherefore, if a straight line, Sec. Q^ E. D. E3 S4 Book H. ' THE ELEMENTS PROP. XI. PROB. ■»46. 1. »'10. 1. *3. 1, " (5. .2. *47. 1. X O divide a given straight line into two parts, so that the rectangle contained by the whole, and one of the parts, shall be equal to the square of the other part. Let AB be the given straight line ; it is required to divide it into two parts, so that the reftangle contained by the whole, and one of the parts, shall be equal to the square of the other part. Upon AB describe^* the square ABDC ; bisect*" AC in E, and join BE; produce CA to F, and make ""EF equal to EB, and upon AF describe* the square of FGHA ; AB is divided in H, so that the redtangle AB, BH, is equal to the square of AH. Produce GH toK ; because the straight line AC is biseded in E, and produced to the point F, the redlangle, CF, FA, to- gether with the square of AE, is equal '^ to the square of EF: But EF is equal to EB j therefore the redangle CF, FA, to- gether with the square of AE, is equal to the square cf EB : and the squares of BA, AE,are equal*^ to the square of EB, because the an- gle EAB is a right angle ; therefore the reftangle CF, FA, together with the square of AE, is equal to the squares of BA, AE : Take away the square of AE, which is common to both, there- fore the remaining redlangle CF, FA, is equal to the square of AB j and the figure FK is the re£langle contained by CF, FA, for AF is equal to FG ; arid AD is the square of AB ; there- fore FK is equal to AD : Take away the common part AK, and the remain- der FH is equal to the remainder HD : And HD is the redlanele contained by AB, BH, for AB is equal to BD j and FH is the square of AH. Therefore the redangle AB, BH is equal to the square of AH : Wherefore, the straight line AB is divided in H, so that the redlangle AB, BH, is equal to the square of AH. Which was tQ be done. OF EUCLID. 55 Book II. PROP. XII. THEOR. IN obtuse angled triangles, if a perpendicular be drawn from any of the acute angles to the opposite side produced, the square of the side subtending the obtuse angle is greater than the squares of the sides containing the obtuse angle, by twice the rectangle, contained by the side upon which, when produced, the perpendicular falls, and the straight line inter- cepted without the triangle between the perpendi- cular and the obtuse angle. Let ABC be an obtuse angled triangle, having the obtuse angle ACB, and from the point A let AD be drawn* perpen- » 12. i. dicular to BC produced : The square of AB is greater thJui the squares of AC, CB, by twice the rectangle BC, CD. Because the straight line BD is divided into two parts in the point C, the square of BD is equal '^o the squares of BC, CD, and A '*'2- twice the reftangle BC, CD : To each of these equals add the square of. DA; and the squares of BD, DA, are equal to the squares of BC, CD, DA, and twice the rectangle BC, CD : But the square of BA is equal •^to the squares of BD, DA, be- ^y^ / j c 47. 1, cause the angle at D is a right B C angle j and the square of CA is equal= to the squares of CD, DA : Therefore the square of BA is equal to the squares of BC, CA, and twice the rectan- gle BC, CD ; that is, the square of BA is greater than the squares of BC, CA, by twice the redangle BC, CD. There- fore, in obtuse angled triangles, &c. Q. E. D. E 4. THE EL.EM ENTS Book II. SeeN, 12. 1. "7. 2. 47. 1. * 16. 1. 1'2. 2. PROP. XIII. THEOR. 1 N every triangle, the square of the side subtend- ing any of the acute angles, is less than the squares of the sides containing that angle, by twice the rectangle contained by either of these sides, and the stra:ight lipe intercepted between the perpendi- cular let fall upon it from the opposite angle, and the acute angle; Let ABC be any triangle, and the angle at B one of its acute angles, and upon BC, one of the sides containing it, let fall the perpendicular* AD from the opposite angle : The square of AC, opposite to the angle B, is less than the squares of CB, BA, by twice the reaangle CB, BD. ' First, Let AD fail within the triangle ABC ; and because the straight line CB is divided into two parts in the point D, the squares of CB, BD are equal** to twice the redtangle contained by CB, BD, and the square of DC : To each of these equals add the square of AD ; therefore the squares ot CB, BD, DA, are equal to twice the redtangle CB, BD ; and the squares ot AD, DC : But the squares of AB is equal= to the square BD, DA, because the angle BDA is a right angle ; and the square of AC is equal to the squares of AD, DC : 'rherefoi:e the squares of CB, BA are equal tp the square of AC, and twice the recStangle CB, BD, that is, the square of AC alone is less than the squares of CB, BA \fy twice the rcdangle CB, BD. Secondly, Let AD fall without the triangle ABC : Then, because the angle at D is a right angle, the angle ACB is greater "^ than a right angle; and therefore the square of AB is equal *= to the squares of AC, CB, and twice the redangle BC,CD : To these equals add the square of BC, and the square OF EUCLID. 57 squares of AB, BC are equal to the square of AC, and twice ^^^kAL the sQuare of BC, and twice the redlangle BC, CD : But be- ^"^ cause BD is divided into two parts in C, the rectangle DB, BC is equaK to the re*Stangie.BC, CD and the square of BC : '3. 2- Arid the doubles of these are equal : Therefore the squares of AB, BC are equal to the square of AC, and twice the rect- angle DB, BC : Therefore the square of AC alone is less ihan the squares of AB, BC by twice the rectangle DB, BC. Lastly, let the side AC be perpendicular to a BC ; then is BC the straight line between the perpendicular and the acute angle at B ; and it is manifest, that the squares of AB, BC, are equals to the square of AC and twice the square of BC : Therefore, in every uiangle, &c. Q. E. D. «47. 1. PROP. XIV. PROB. X O describe a square that shall be equal to a given s<«^. rectilineal figure. Let A be the given redilineal figure ; it is required to de- scribe a square that shall be equal to A. Describe* the redtangular parallelogram BCDE equal to the ^^^ j redtilineal figure A. If then the sides of it BE, ED are equal to one another, it is a square, and what was required is now done : But if they are not equal, produce one of them BE to F, and make EF equal to ED and bisect BF in G : and from the centre G, at the distance GB, or GF, describe the semi- circle BHF, and produce DE to H, and join GH : Therefore because the straight line BF is divided into two equal parts ia the point G, and into two unequal at E, the re<^ngle BE, EF, together with the square gf EG, is equal^ to the square of ^' *• GF : But GF is equal to GH : therefore the redangle BE, EF, 58 THE ELEMENTS, &c. BookII. together with the square of EG, is equal to the square of GH : T^^""^/ But the squares of HE, EG are equal*^ to the square of GH : ''/■ ' Therefore the re6tangle BE, EF, together with the square of EG, is equal to the squares of HE, EG : Take away the square of EG, which is common to both ; and the remaining redlangle BE, EF is equal to the square of EH : But the redlangle contained by BE, EF is the parallelogram BD, be- cause EF is equal to ED ; therefore BD is equal to the square* of EH ; but BD is equal to the re<Slilineal figure A ; there- fore the redlilineal figure A is equal to the square of EH. Wherefore a square has been made equal to the given re<5lili- neal figure A, viz. the square described upon EH. Which was to be done. THL 79 ELEMENTS OF EUCLID. BOOK. III. DEFINITIONS. I. IliQUAL circles are those of which the diameters are "equal, <>r from the centres of which the straight lines to the circunv- ferences are equal. < This is not a definition, but a theorem, the truth of which * is evident ; for, if the circles be applied to one another, so * that their centres coincide, the circles must likewise Coin- * cide, since the straight lines from the centres are equal.' II. A straight line is said to touch a circle, when it meets the circle, and being produced does not cut it. III. Circles are said to touch one another, which meet but do not cut one another. IV. Straight lines are said to be equally dis- tant from the centre of a circle, when the perpendiculars drawn to them from the centre are equal. V. And the straight line on which the gr«ater perpendicular falls, is said to be farther from the centre. 6(3 Book III, THE ELEMENTS VI. A segment of a circle is the figure con- tained by a straight line and the cir- cumference it cuts ofF. VII. *' The angle of a segment is that which is contained by the " straight line and the circumference." VIII. An angle in a segment is the angle con- tained by two straight lines drawn from any point in the circumference of the segment, to the extremities of the straight line which is the base of the segment. IX. And an angle is said to insist or stand upon the circumference intercepted between the straight lines that contain the angle. X. The sector of a circle is the figure con-* tained by two straight lines drawn from the centre, and the circumference be- tween them. XI. Similar segments of a circle are those in which the an- gles are equal, or which contain equal angles. PROP. I. PROB. N- X O find the centre of a given circle. Let ABC be the given circle; it is, required to find itj centre. Draw within it any straight line AB, and bise£l* it in D ; from the point D draw '' DC at right angles to AB, and pro- duce it to E, and biseft CE.in F : The point F is the centre of the circle ABC. Foi » 10. 1. ^ U. 1. OF EUCLID. 6l For, if it be not, let, if possible, G be the centre, and join BooillL GA, GD, GB : Then, because DA is equal to DB, and DG ^•*^^'*^ common to the two triangles ADG, BDG, the two sides AD, DG are equal to the two BD, DG, each to each ; and the base GA is equal to the base GJ3, because they are drawn from the centre G* : Therefore the angle ADG is equal ' to the angle GDB : But when a straight line standing upon another straight line makes the adjacent angles equal to one another, each of the angles is a right angle"* : Therefore the angle GDB is a right angle: But FDB is likewibe a right angle; wherefore the angle FDB is equal to the anale GDB, the greater to the less, which is impos- sible : Therefore G is not the centre of the circle ABC . In the same manner it can be shewn, that no other point but F is the centre : that is, F is the centre of the circle ABC ; Which was to be found. ;' Cor. From cnis it is manifest, that if in. a circle a straight line bisect another at right angles, the centre of the circle is in the line which biseds the other. <10Def;i. PROP. II. THEOR. 1 F any two points be taken in the circumference of a circle, the straight line which joins them shall fall withhi the circle. Let ABC be a circle, and A, B any two points in thecircum> ference ; the straight line drawn from A to B shall fall within the circle. For, ifitdo not, let it fall, if possible, without, as AEB j find * D the centre of the circle ABCj and join AD, DB, and produce DF, any straight line meeting the circumference AB to E : Then because DA is equal to DB, the angle DAB is equal*" to the angle DBA ; and because AE, a side of the triangle ■♦ N. B. Wbtaevcr the expressfba " straight lines from the centre," or " dra'wn from the e«nue," occurs, it is to b« understood that thej' are drava to the drcnm- fcrcnce, DAE, 1.3. *«. 1. 62 THEELEMENTS ^ooK III. DAE, is produced to B, the angle DEB is greater "^ than the j'*?r^p^ angle DAE ; but DAE is equal to the angle DBE ; therefore the angle DEB is greater than the angle DBE : But to the *,19. 1, greater angle the greater side is opposite"^ ; DB is therefore greater than DE : But DB is equal to DF i wherefore DF is greater than DE, the less than the greater, which is impos- sible : Therefore the straight line drawn from A to B does not fall without the circle. In the same manner, it may be demonstrated that it does not fall upon the circumference ; it falls therefore within it. Wherefore, if any two points, &c, Q^E. D. PROP. III. THEOR. xF a straight line drawn through the centre of a circle hisect a straight line in it which does not pass through the centre, it shall cut it at right angles ; and if it cuts it at right angles, it shall bisect it. Let ABC be a circle; and let CD, a straight line drawn through the centre, bisedi any straight line AB, which does not pass through the centre, in the point F : It cuts it also at right angles. * '• ^' Take^ E the centre of the circle, and join EA, EB. Then, because AF is equal to FB, and FE common to the two tri- angles AFE, BFE, there are two sides in the one equal to two sides in the other, and the base EA is equal to the base EB j therefore the * 8. 1. angle AFE is equaP to the angle BFE: But when a straight line standing upon another niakes the adjacent angles tf qual to one another, each of them is a right 'iODef.i. «^angle : Therefore each of the angles AFE, BFE is a right angle ; wherefore the straight line CD„ drawn through the centre bisecting another AB that does not pass through the centre, cuts the same at right angles. But let CD cut AB at right angles ; CD also bisects it, that is, AF is equal to FB. The same construction being made, because EA, EB from <5. 1. ^l'® centre are equal to one another, the angle EAF is equal** to the angle EBF : and the right angle AFE is equal to the right angle BFE : Therefore, in the two triangles, EAF, 2 EBF^ OF EUCLID. 6,3 EBF, there are two angles in one equal to two angles in the Booi in . other, and the side EF, which is opposite to one of the equal ^^"^'^^ angles in each, is common to both ; therefore the other sides • 26. 1. are equal'^i AF therefore is equal to FB. Wherefore, if a straight line, &c. Q^ E. D. PPOP. IV. THEOR. J.F in a circle two straight lines cut one another which do not both pass through the centre, they do not bisect each other. Let ABCD be a circle, and AC, BD two straight lines In it which cut one another in the point E, and do not both pass through the centre : AC, BD do not bisecft one another. For, if it is possible, let AE be equal to EC, and BE to ED: If one of the lines pass through the centre, it is plain that it cannot be bisected by the other which does not pass through the centre ; But if neither of them pass through the centre, take * F the centre of the cir- cle, and join EF : and because FE, a straight line through the centre, bi- sefls another AC which does not pass through the centre, it shall cut it at right'' angles J wherefore FEA is a right angle : Again, because the straight line FE bisedls the straight line BD which does not pass through the centre, it shall cut it at right'' angles ; where- fore FEB is a right angle : And FEA was shewn to be a right angle ; therefore FEA is equal to the angle FEB, the less to the greater, which is impossible : Therefore AC, BD do not biseS one another. Wherefore, if in a circle, &c. Q. E. D. PROP. V. THEOR. JlF two circles cut one another, they shall not have the same centre. Let the two circles ABC, CDG cut one another in the |)oints B, C i they have not tl^ same centre. For 64 Book III THE ELEMENTS For, if it be possible, let E be their centre ; join draw any straight line EFG meet- ing them in F and G ; and because E is the centre of the circle ABC, CE is equal to EF : Again, be- cause E is the centre of the circle CDG, CE is equal to EG : But CE was shewn to be equal to EF ; therefore EF is equal to EG, the less to the greater, which is im- possible : Therefore E is not the centre of the circles ABC, CDG. Wherefore, if two circles, &c. Q. E. D. EC, and PROP. VI. THEOR. JlF two circles touch one another internally, they shall not hav^e the same centre. Let the two circles ABC, CDE, touch one another inter- nally in the point C : They have not the same centre. For, if they can, let it be F; join FC and draw any straight lineFEB meeting them in E and B J And because F is the centre of the circle ABC, CF is equal to FB ; Also, because F is the centre of the circle CDE, CF is equal to FE : And CF was shewn equal to FB ; therefore FE is equal to FB, the less to the greater, which is impossible : V/herefore F is' not the centre of the circles ABC, CDE. Therefore, if two circles, &c. Q^ E. D. OF EUCLID. 65 Book III. PROP. VII. THEOR. • — < — XF any point be taken in the diameter of a circle Avhich is not the centre, of all the straight lines which can be drawn froin it to the circumference, the greatest is that in which the centre is, and the other part of that diameter is the least; and, of any others, that which is nearer to the line which passes through the centre is always greater than one more remote : And from the same point there can be drawn only two straight lines that are equal to one another, one upon each side of the shortest line. Let ABCD be a circle, and AD its diameter, in which let any point F be taken which is not the centre : Let the centre be E ; of ail the straight lines FB, FC, FG, &c. that can be drawn from F to the circumference, FA is the greatest, and ! FD, the other part of the diameter BD, is the least : And of the others, FB is greater than FC, and FC than FG. Join BE, CE, GE ; and because two sides of a triangle are greater^* than the" third, BE, EF are greater than BFj but AE *20. l. , is equal toEB ; therefore AE, tF, that is AF,is greater than BF: A- gain, because BE is equal to CE, and FE common to the triangles BEF, CEF, the twosidesBE,£F are equal to the two CE, EF i but the angle BEF is greater than the jangleCEF; therefore the base BF \ /^jiF\ / to^. 1. I is greater*" than the base FC : For ! the same reason, CF is greater than ' GF : Again, because GF, FE are ' greater^ than EG, and EG is equal to ED ; GF, FE are greater than ED : Take away the com- mon part FE, and the remainder GF is greater than the re- mainder FD : Therefore FA is the greatest, and FD the least of all the straight lines from F to the circumference i and BF is greater than CF, and CF thaii GF. Also there can be drawn only two equal straight lines from the point F to the circumference, one upon each side of the F shortest 66 THEELEMENTS Book III. shortest line FD : At the point E in the straight line EF, ^^''^ make'' the angle FEH equal to the angle GEF, and join FH : Then because GE is equal to EH, and £F common to the two triangles GEF, HEF ; the two sides GE, EF are equal to thjC two HE, EF j and the angle GEF is equal to the an- '^ -i. 1. gle HEF ; therefore the base FG is equal"* to the base FH : 5ut, besides FH, no other straight line can be drawn from F to the circumference equal to FG : For, if there can, let it be FK ; and because FK is equal to FG, and FG to FH, FK is equal to FH ; that is, a line nearertothat which passes through the centre, is equal to one which is more remote ; which is impossible. Therefore, if any point be taken, &c. Q. E. D. PROP. VIII. THEOR. If any point be taken without a circle, and straight lines be drawn from it to the circumference, where- of one passes through the centre; of those which fall upon the concave circumference, the greatest is that which passes through the centre ; and of the rest, that, which i§ nearer to that through the centre is al- ways greater than the more remote : But of those which fall upon the convex circumference, the least is that between the point without the circle and the diameter ; and of the rest, that which is nearer to the least is always less than the more remote : And only two equal straight lines can be drawn from the point into the circumference, one upon eacli side of the least. Let ABCbeacircle,and Danypoint without it, from which let the. straight lines DA, DE, DF,.DC be drawn to the cir- cumference, whereof DA passes through the centre. Of thos* which fall upon the concave part of the circumference AEFC, tiie greatest is AD which passes through the centre ; and the nearer to it is always greater than the more remote, viz. DE than DF, and DF than DC: But of those which fall upon the eonvcx circumference HLKG, the least is DG between the* I point I At. OF EUCLID. . 6f Book 111. Jjolnt D and the diameter AG ; and the nearer to it is always ^--v**-', less than the more remote, viz. DK than DL, and DL than DH. Take* M the centre of the circle ABC, and join ME,MF, ' i- 3. MC, MK, ML, MH : And because AM is equal to ME, add MD to each, therefore AD is equal to EM, MD i but EM,MD are greater^ than ED ; therefore also AD is greater than ED " 20. i. Again, because ME is equal to MF, and MD common to the triangles EMD, FMD ; EM, MD are equal to FM, MD ; but the angle EMD is greater than the angle FMDj therefore the base ED is greater'^ than the base FD : In like manner it may be shewn that FD is greater than CD : Therefore DA is the greatest ; and DE greater than DF, and DF than DC : And because MK, KD are greater'' than MD, and MK is equal to MG, the remainder KD ' is greater ^ than the remainder ! GD, that is GD is less than KD : ■ And because MK, DK are drawn to the point K within the triangle MLD from M, D, the extremi- ties of its side MD, MK, KD are Jess = than ML, LD, whereof MK is equal to ML j therefore the remainder DK is less than the remainder DL : In like manner it may be shewn, that DL is less than DH : Therefore DG is the least, and DK less than DL, and DL than DH : Also there can be drawn only two equal straight lines from the point D to the circumference, j one upon each side of the least : At the point M, in the straight line MD, make the angle DMB equal to the angle DMK, land join DB : And because MK is equal to MB, and MD I common to the triangles KMD, BMD, the two sides KM, |MD are equal to the two BM, MD ; and the angle KMD lis equal to the angle BMD ; therefore the base DK is equaK ^ *• ^• to the base DB : But, besides DB, there can be no straight jline drawn from D to the circumference equal to DK : For, jif there can, let it be DN j and because DK is equal to DN, iand also to DB ; therefore DB is equal to DN, that is, the ,nearer to the least equal to the more remote, which is im- ipossible. If therefore, any point, &c. Q^ E. D, 1 F2 ■ ' 21.1. 68 Book III. THE ELEMENTS PROP. IX. THEOR. »7. J[ F a point be taken within a circle, from which there fall more than two equal straight lines to the cir- cumference, that point is the centre of the circle. Let the point D be taken within the circle ABC, from which to the circumference there fall more than two equal straight lines, viz. DA, DB, DC, the point D is the centre of the circle. For, if not, let E be the centre, join DE and produce it- to the cir- cumference in F, G : then FG is a diameter of the circle ABC : And because in FG, the diameter of the circle ABC, there is taken the point D, which is not the centre, DG shall be the greatest line from it to the circumference, and DC great- er » than DB, and DB than DA : But thej^ are likewise equal, which is impossible ; Therefore E is not the centre of the circle ABC: In like manner, it may be demon- strated, that no other point but D is the centre ; D therefore is the centre. Wherefore, if a point be taken, &c. Q. E. D, PROP. X. THEOR. i^NE circumference of a circle cannot cut anothe in more than two, points. If it be possible, let the circumfe- rence FAB cut the circumference DEF in more than two points, viz. in. B, G, F ; take the centre K of the ^ cirrk- ABC, and join KB, KG, KF: 1^ And because within the circle DEF there is taken the point K, from which to the circumference DEF fall more tlian two equal straight lines KB, KG, KF, the point K is'" OF EUCLID."- 6^ the centre of the circle DEF : But K is also the centre of the Boo* iii. circle ABC; therefore the same point is the centre of two cir- cles that cut one another, which is impossible^ Therefore " 5. C\ one circumference of a circle cannot cut another in more than ^wo points. Q_ E. D. PROP. XI. THEOR. X F two circles touch each other internally, the straight line which joins their centres heing pro- iduced shall pass through the point of contact. Let the two circles ABC, ADE touch each other internally in the point A, and let F be the centre of the circle ABC, ar.d G the centre of the circle ADE : The straight line which joins the centres F, G, being producedj passes through the point A. For, if not, let it fall otherwise, if possible, as FGDH, and join AF, AG: And because AG, GF are great- ec * than FA, that is, than FH, for \ \^ ^ ^^^^^ ' *°' ^' FA is equal to FH, both being from the same centre ; take away the com- mon part FG; therefore the remain- der AG is greater than the remainder GH; But AG is equal to GD ; therefore GD is greater than GH, the less than the greater, which is impossible. Therefore the straight line whichjoins the points F, G cannot fall otherwise than upon the point A, that is, it must pass through it. Therefore, if two circles, Sec. Q^ E. D. PPOP. XII. THEOR. J.F two circles touch each other externally, the straight line which joins their centres shall pass through the point of contact. Let the two circles ABC, ADE, touch each other exter- nally in the point A ; and let F be the centre of the circle ABC, and G the centre of ADE : The straight line which joins the points F, G shall pass through the point of contact A. For, if not, let it pass otherwise, if possible, as FCDG, and F 3 join 7« THE ELEMENTS B«ojc in. join FA, AG: And because F is the centre of the circls ABa ^^''"^ AF is equal to FC : Also because G is thecentreof ' the circle ADE, AG is equal to GD: Therefore FA, AG are equal to FCj DG i wherefore the whole FG is greater than FA, AG : But it is also •20. I, less*; which is impossi- ble : Therefore the straight line which joins the points F, G shall not pass otherwise than through the point of contact A, that is, it must pass through it. Therefore, if two circles, &c. Q. E. D. PROP. XIII. THEOR. vJNE circle cannot touch another in more points than one, whether it touches it on the inside or outside. For, if it be possible, let the circle EBF towch the circle ABC in more points than one, and first on the inside, in the « 10.11, 1. points B, D ; join BD, and draw* GH bisecting BD at right angle.s ^.Therefore because the points B, D are in the circum- H B "1.3. •^Cor. 1.3, " 11, 3. ference of each of the circles, the straight line BD falls within each'' of them : And their centres are<^ in the straight line GH which bisedts BD at right angles : Therefore GH passes through the point oFcontacf*; but it does not pass through it, because the points B, D are without the straight line GH, which is absurd : Therefore one circle cannot touch another on the inside in more points than one. Nor can two circles* touch one another on the outside in more OF EUCLID. 7» more than one point ; For, if it be possible, letthecircle ACK Book jH. touch the circle ABC in the points A, C, and join AC : There- ^'^^''^^ fore, because the two points A, C are in the circumference of the circle ACK, the straight line AC which joins them shall fall within** the circle ACK : And the circle ACK is without the circle ABC J and therefore the straight line AC is without this last circle ; but be- cause the points A, C are in the circum- ference of the circle ABC, the straight line AC must be within'' the same cir- cle, which is absurd : Therefore one circle cannot touch another on the out- side in more than one point : And it has been shewn, that they cannot touch on the inside in more points than one. Therefore, one circle, &c. Q. E. D. PROP. XIV. THEOR. JCuQUAL straight lines in a circle are equally dis- tant from the centre; and those which are equally distant from the centre, are equal to one another. Let the straight lines AB, CD, in the circle ABDC, be e- qual to one another j they are equally distant from the centre. TakeE the centre of the circle ABDC, and from it draw EF, EG perpendiculars to AB, CD : Then, because the straight line EF, passing through the centre, cuts thsstraight line AB, which does not pass through the centre, at right angles, it also bisefls' it : Where- fore AF is equal toFB, and AB double of AF. For the same reason CD is double of CG : And AB is equal to CD ; therefore AF is equal to CG ; And because AE is equal to EC, the square of AE is equal to the square of EC -. But the squares of AF, FE are equal'' to the square of AE, because the angle AFE is a right angle; and for the like reason, the squares of EG, GC are equal to the square of EC : Therefore the squares of AF, FE are equal to the squares" of CG, GE, of which the square of AF is equal to F 4 the 72 THE ELEMENTS ^°^*I"- the square of CG i because AF is equal to CG ; therefore the remaining square of FE is equal to the remaining square of EG, and the straight line EF is therefore equal to EG : But straight lines in a circle are said to be equally distant from the centre, when the perpendiculars drawn to them from the cen- •4.def, 3. tre areequal'^: Therefore AB, CD be equally distant from the centre. Next, if the straight lines AB, CD be equally distant from the centre, that is if FE be equal to EG; AB is equal to CD: For, the same construction being made, it may, as before, be demonstrated, that AB is double of AF, and CD double of CG, and that the squares of EF, FA are equal to the squares of EG, GC ; of which the square of FE is equal to thesquarc of EG, because FE is equal to EG ; therefore the remaining square of AF is equal to the remaining square of CG; and the Straight line AF is therefore equal to CG: And AB is double of AF, and CD double of CG ; wherefore AB is equal to CD. Therefore equal straight lines, &c. Q^ E. D. PROP. XV. THEOR. SeeN. 10. I. 1 HE diameter is the greatest straight line in a cir- cle ; and, of all others, that which is nearer to the centre is always greater than one more remote : and the greater is nearer to the centre than the less. Let ABCD be a circle, of which the diameter fs AD, and the centre E ; and let BC be nearer to the centre than FG ; AD is greater than any straight line BC which is not a dia- meter, and BC greater than FG. Fi-om the centre draw EH, EK per- pendiculars to BC, FG, and join EB, EC, EF ; and because AE is equal to EB, and ED to EC, AD is equal to EB, EC ; but EB, EC are greater •' than BC : wherefore, also AD is greater than BC. Apd, because BC is nearer to the cehtre than FG, EH Is less OF EUCLID. 73 less^ than EK . But, as was demonstrated in the preceding. Book IH. BC is double of BH, and FG double of FK, and the squares of e ^ ^^ 3^ EH, HB are equal to the squares of EK, KF, of which the square of EH is less than the square of EK, because EH is less than EK ; therefore the square of BH is greater than the square of FK, and the straight line BH greater than FK. and therefore BC is greater than FG. Next, Let BC be greater than FG j BC is nearer to the cen- tre than FG, that is, the same construction being made, EH is less than EK : Because BC is greaterthan FG, BH likewise is greater than KF : And the squares of BH, HE are equal to the squares of FK, KE, of which the square of BH is greater than the square of FK, because BH is greater than FK ; therefore the square of EH is less than thesquare of EK, and the straight \ lilieEHless than EK. Wherefore the diameter, &c, Q^E. D. PROP. XVI. THEOR. JL HE straight line drawn at right angles to thedia- seeN. meter of a circle, from the extremity of it, falls without the circle; and no straight linecanhe drawn between that straightlineand the circumference from the extremity, so as not to cut the circle ; or, which is the same thing, no straight line can make so great an acute angle with the diameter at its extremity, or so small an angle with the straight line, which is at right angles to it, as not to cut the circle. Let ABC be a circle; the centre of which is D, and the dia- meter AB : the straight line drawn at right angles to AB from its extremity A, shall fallwithout the circle. For, if it does not, let it fall, if possible, within the circle, as AC, and draw DC to the point C where it meets the circumference : And because DA is equal to DC, the angle DAC is equal^ to the angle ACD; but DAC is a right angle, therefore ACD is a right angle, and the angles DAC, ACD are therefore equal to two right angles j which is impossible ^ : Therefore u THE ELEMENTS ' 12. 1. * i?. 1, Book iu. Therefore the straight line drawn from A at right angles to. ^•^'^^'^^ j8A does not fall within the circle: In the same manner, it may be demonstrated, that it does not fall upon the circum- ference; therefore it must fall without the circle, as AE. And between the straight line AE and the circumference no straight line can be drawn from the point A which -does not cut the circle : For, if possible, let FA be between them, and from the point D draw '^ DG perpendicular to FA, and let it meet the circumference in H : And because AGO is a right angle, and DAG less ^ than aright angle, DA is greater'^ than DG: But DA is equal to DH : therefore DH is greater than DG, the less than the greater, which is iitipossible : Therefor'^ no straight line can be drawn from the point A between AE and the circumfe- rence, which does not cut the cir- cle, or, which amounts to the same thing, however great an acute angle a straight line makes with the dia- meter at the point A, or however small an angle it makes with AE, the circumference passes between that straight line and thff perpendicular AE. * And this is all that is to be understood, * when, in the Greek text, and translations from it, the angld * of the semicircle is said to be greater than any atute reitili-.. * ncal angle, and the remaining angle less than any rectilineal * angle.' Q. E. D. Cor. From this it is manifest, that the straight line which is drawn at right angles to the diameter of a circle from the ex-, tremity of it, touches the circle ; and that it touches it only in one point, because, if it did meet the cifcle in two, it would • 2. 3. fall within it' . * Also it is evident that there can be but one ' straight line which teuches the circle in the same point. * PROP. XVII. PROB. l O draw a stiaight line from a given point, either .witkout or iu the circumference, which sl^ll touch a given circle. First, let A be a given point without the given circle BCDj OF EUCLID. 75 » 1.3, xil. 1. it is required to draw a straight line from A which shall touch ^^^^^' the circle. Find * the centre E of the circle, and join AE ; and from the centre E, at the distance EA, describe the circle AFGj from the point D draw '' DF at right angles to EA, and join EBF, AB. AB touches the circle BCD. Because E is the centre of the circles BCD, AFG, EA is equal to EF : And ED to EB ; therefore the two sides AE, EB are equal to the two FE, ED, and they contain the angle at E common to the two tri- angles AEB, FED ; there- fore the base DF is equal to the base AB ; and the triangle FED to the trian- gle AEB, and the other angles to the other angles'^ : There- «4. i. fore the angle EBA is equal to the angle EDF : But EDF is a rightanglcjWhereforeEBA.isaright angle: And EB is drawn from the centre : But a straight line drawn from the extremi- ty of a diameter, at right angles to it, touches the circle<* : *""' ' Therefore AB touches the circle ; and it is drawn from the given point A. Which was to be done. But if the given point be in the circumference of the circle, as the point D, draw DE to the centre E, and DF at right angles to DE ; DF touches the circle"^. PROP. XVIII. THEOR. JLF a straight line touches a chcle, the straight line drawn from the centre to the point of contact, shall be pei-pendicular to the line touching the circle. Let the straight line DE touch the circle ABC in the point Ci take the centre F, and draw the straight line FC : FC is perpendicular to DE. For, if it be not, from the point F draw FBG perpendicular to D£ J and because FGC is a right angle, GCF is** an acute b 17 j angle i and to the greater angle the greatest"^ side is opposite : « 19! \. Therefore 76 THE ELEMENTS ^^^2^^2!L' Therefore FC is greater than FG ; ^^"^^^^^ but FC is equal to FB ; therefore FB is greater than FG, the less than the greater, which is impos- sible : Wherefore FG is not per- pendicular to DE : In the same manner it may be shewn, that no other is perpendicular to . it besides FC, that is FC is perpendicular to DE» Therefore, if a straight line, &c. Q- E. D. G E PROP. XIX. THEOR. If a straight line touches a circle, and from thr point of contact a straight line be drawn at right angles to the touching Une, the centre of the circle shall be in that line. Let the straight line DE touch the circle ABC In C, and from C let CA be drawn at right angles to DE j the centre »f the circle is in CA. For, if not, let F be the centre, if possible, and join CF j Because DE touches thecircle ABC, and FC is draw-n from the centre to the point of contact, FC is perpendi- ' IS. 3. cular * to DE ; therefore FCE is a right angle : But ACE is also a right angle ; therefore the angle FCE is equal to the angle ACE, the less to t^\^ the greater, which is impossible : Wherefore F is not the centre of the circle ABC: In the same manner, it may be shewn, that no 'other point which it not in CA, is the centre ; that is, the centre is in C A , Therefore, if a straight line, &c. Q. E. D. prop'. XX. THEOR. ',2. N. 1 HE angle at the centre of a circle rs double of the angle at the circumference, upon the same base, that is, upon the same part of the circumference Let OF EUCLID. 77 »5. 1. *3^1. Let ABC be a circle, and BEC an angle at the centre, and Book ill. BAC an angle at the circuniference, which have the same cir- ^■'•**''*^ cumference BC for their base •, the angle BEC IS double of the angle BAC. First, let E, the centre of the circle, be within the angle BAC, and join AE, and produce it to F : Because EA is equal to EB, the angle EAB is equal'' to the angle EEA; therefore the angles EAB, EBA are double of the angle EAB j but the angle BEF is equal'' to the angles EAB, EBA ; therefore also the angle BEF is double of the angle EAB : For the same reason, the angle FEC is double of the angle E AC : Therefore the whole angle BEC is doubleof the whole angle BAC. Again, let E, the centre of the circle, be without the angle BDC, and join DE, and produce it to G. It may be demonstrated, as in the first case, that the angle GEC is double of the angle GDC, and that GEB, a part of the first, is double of GDB, a part »f the other ; therefore the re- Gr maining angle BEC is double of the remaining angle BDC. Therefore the angle at the centre, &c. Q^E. D. PROP. XXI. THEOR. JL HE angles in the same segment of a circle are equal to one another. Let ABCD be a circle, and BAD, BED angles in the same segment BAED : l^he angles BAD, BED are equal to one another. Take F, the centre of the circle ABCD: And, first, let the segment. BAED be greater than a semicircle, and join BF, FD : And because the angle BFD is at the centre, and the angle BAD at the circumference, a;)d that they have the same part of the SesK. Book III, » 20. 3. THE ELEMENTS the circumference, viz. BCD for their base ; therefore the an- gle BFD is double* of the angle BAD : For the same reason, the angle BFD is double of the angle BED: Therefore the angle BAD is equal to the angle BED. But, if the segment BAED be not greater than a semicircle, let BAD, BED be angles in it ; these . -j, also are equal to one another : Draw Jk-^j^ AF to the centre, and produce it to C, and join CE : Therefore the seg- "A ment BADC is greater than a semi- circle ; and the angles in it BAG, BEC are equal, by the first case : For the same reason, because CBED is greater than a semicircle, the angles CAD, CED are equal : Therefore the whole angle BAD is equal to the whole angle BED. Wherefore the angles in the same seg- ment, &c. Q. E. D. PROP. XXII. THEOR. •32. 1. *21.: X HE opposite angles of any quadrilateral figure described in a circle, are together equal to two right angles. Let ABCD be a quadrilateral figure in the circle ABCD ; any two of its opposite angles are together equal to two right angles. Join AC, BD ; and because the three angles of every tri- angle are equal* to two right apgles, the three angles of the triangle CAB, viz. the angles CAB, ABC, BCA are equal to two right angles : But the angle CAB j) is equal^ to the angle CDB, because v'-'''7"^^"<:>v r they are in the same segment BADC, /^ / \ ^'^'-' and the angle ACB is equal to the angle ADB, because they are in the same segment ADCB : Therefore the whole angle ADC is equal to the an- "\ /B gle CAB, ACB: To each of these equals add the angle ABC ; therefore the angles ABC, CAB, BCA are equal to the angles ABC, ADC : But ABC, CAB, BCA are equal to two right angles ; therefore alsothe angles ARC, ADC are equal to two right angles : In the same manner, the angles BAD OF EUCLID. 79 BAD, DCB, may be shewn to be equal to two right angles. Boo» in. Therefore, the opposite angles, &c. Q^ E. D. >*'n'^«' PROP. XXIII. THEOR. U PON the same straight line, and upon the same see k. side of it, there cannot be two similar segments of circles, not coinciding with one anotiier. If it be possible, let the twe similar segments of circles, viz. ACB, ADB, be upon the same side of the same straight line AB, not coinciding with one another: Then, because the cir- cle ACB cuts the circle ADB in the two points A, B, they cannot cut one another in any other point* : One of the segments must therefore fall within the other : Lee ACB fall within ADB, and draw the straight line BCD, and join CA, DA : And because the seg- ment ACB is similar to the segment ADB, and that similar segments of circles contain'' equal angles; the angles ACB is ^iLdef-S. equal to the angle ADB, the exterior to the interior, which is impossible.*^ rherefore, there cannot be two similar seg^ <= J6. i. ments of a circle upon the same side of the same line, which do not coincide. Q^ E. D. • 10. 3. PROP. XXIV. THEOR. OlMILAR segments of circles upon equal straight Sm n. lines, are equal to one another. Let AEB, CFD be similar segments of cricles.upon-the equal straight lines AB, CD j the segment AEB is equal t# the segment CFD, For if the seg- ment AEB be applied to the segments CFD, so as the point A be on C, and A B C ~ D the straight line AB upon CD, the point B shall x:oincide with the point 9, be- cause 8o THE ELEMENTS Book in, cauSc AB is equal te CD ; Therefore the straight line AB co- •^23?3/ inciding with CD, the segment AEB must ^ coincide with the segment CFD, and therefore is equal to it. Wherefore similar segments, &cc. Q. E. D. SeeN. «10. 1. 6. 1. "9.3. PROP. XXV. PROB. A SEGMENT of a circle beinsr oiven, to describe '» &■ the circle of M^hich it is the segment. Let ABC be the given segment of a circle ; it is required to describe the circle of which it is the segment. Bised* AC in D, and from the point D draw** DB at right angles to AC, and join AB : First, let the angles ABD, BAD be equal to one another ; then the straight line BD is equal*^ to DA, and therefore to DC ; and because the three straight lines DA, DB, DC, are all equal j D is the centre of the cir-^ cle.'' From the centre D, at the distance of any of the three DA,DB,DC,describe a circle ; this shall pass through the other points i and the circle of which ABC is a segment is described j And because the centre D is in AC, the segment ABC is a se- 23. 1. ■*. 1. micircle : But if the angles ABD, BAD are not equal to one another,at the point A, in the straight lineAB make'= the angle BAE equal to the angle ABD, and produce BD,if necessary, to E,andjoinEC: And because the angle ABE is equal to the angle BAE, the straight line BE is equal '^ to EA : And because AD is equal to DC, and DE common to the triangles ADE, CDE, the two sides AD, DE are equal to the two CD, DE, each to each J and the angle ADE is equal to the angle CDE, for each of them is a right angle; therefore the base AE is equal '^to the base EC : But AE was shewn to be equal to EB, where- fore also BE is equal to EC : And the three straight lines AE, EB, OF EUCLtD. Si EB, EC are therefore equal to one another ; wherefore*^ E is ^^^- the centre of the circle. From the centre E, at the distance of any of the three AE, EB, EC, describe a circle, this shall pass " 9- 3. through the other points j and the circle of which ABC is a seg- ment is described: And it is evident, that if the angle ABD be greater than the angle BAD, the centre E falls without the segment ABC, which therefore is less than a semicircle : But if the angle ABD be less than BAD, the centre E falls withia the segment ABC, which is therefore greater than a semicircle: Wherefore a segment of a circle being given, the circle is de- scribed of which it is a segment. Which was to be done. PROP. XXVI. THEOR. J. N equal circles, equal angles stand upon equal circumferences, whether they be at the centres or circumferences. Let ABC, DEF be equal circles, and the equal angles BGC, EHF at their Centres, and BAC, EDF at their circumferences: The circumference BKC is equal to the circumierence ELF. Join BC, EF j and because the circles AB J, DEF are equal, the straight lines drawn from their centres are equal : There- fore the two sides BG, GC, are equal to the two EH, HF ; md the angle at G is equal to the angle at H ; therefore the l)ase BC is equal* to the base EF. And because the angle at A » 4. i. s equal to the angle at D, thesegment BAC is similar'' to the ''-^^•«*<.7' Kgment EDF ; and they are upon equal straight lines BC, EF j 3ut similar segments of circles upon eqiaai straight lines are :qual<= to one another, therefore the segment BAC is equal to c ^i. 3. ihe segment EDF: But th? whole circle ABC is equ^l to the G whole 82 THE ELEMENTS Book III. whole DEF , therefore the remaining segment BKC is equal '^^'^^'''^ to the remaining segment ELF, and the circumference BKC to the circumference ELF. Wherefore, in equal circles, Sec. Q. E. D. PROP. XXVn. THEOR. »2. 8. XN equal circles, the angles which stand upon equal circumferences are equal to one another, AV'hether they be at the centres or circumferences. Let the angles BGC, EHF at the centres, and BAG, EDF at the circumferences of the equal circles ABC, DEF stand upon the equal circumferences BC, EF : The angle BGC is equal to the angle EHF, and the angle BAC to the angle EDF. If the angle BGC be equal to the angle EHF, it is manifest * that the angle BAC is also equaj to EDF. But, if not, one 23. 1. * 2». 3. of- them is the greater : Let BGC be the greater, and at the point G, in the straight line BG, make*" the angle BGK equal to the angle EHF; but equal angles stand upon equal circum- ferences% when they are at the centre; therefore the circura-, ference BK is equal to the circumference EF : But EF is equaJ to BC; therefore also BK is equal toBC, the less to the greater, which is impossible: Therefore the angle BGCis not unequal to the angle EHF ; that is, it is equal to it : And the angle at A is half of the angle BGC, and the angle at D half of th( angle EHF : Therefore the angle at A is equal to the angl< at D. Wherefore, in equal circles. &c, C^ E. P. > OF EUCLID. 83 Boon III. PROP. XXVIII. THEOR. In equal circles, equal straight lines cut off equal circumteiences, the greater equal to the greater, and the le88 to the less. Let ABC, DEF be equal circles, and BC, EF equal straight lines in them, which cut ofF the two greater circumferences BAG, EDF, and the two less BGC, EHF : the greater BAC is equal to the greater EDF, and the less BGC to the less EHF. Take^K,L, the centresofthecircles, and join BK,KC,EL, » 1. 2. LF : And because the circles are equal, the straight lines from Cr their centres are equal ; therefore BK, KC are equal to EL, LF ; and the base BC is equal to the base EF ; therefore the angle BKC is equal'' to the angle ELF : But equal angles stand « s. l. upon equal"^ circumferences, when they are at the centres j c ^g. ^ therefore the circumference BGC is equal to the circumfe- rence EHF. But the whole circle ABC is equal to the whole EDF J the remaining part therefore of the circumference, viz. BAC, is equal to the remaining part BDF. Therefore, ia equal circles, &c. Q^ E. D. PROP. XXIX. THEOR. N equal circles, equal circumferences are sub- tended by equal straight lines. Let ABC, DEF be equal circles, and let the circumferences BGC, EHF also be equal j and join BC, EF ; The straight line BC is equal to the straight line EF. G 2 Take 84 THE ELEMENTS Book III. Tak€^ K, L, the centres of the circles, and join BK, KC, ^J'g'^^ EL, LF : And because the circumference BGC is equal to the 27. '^i. 1. circumference EHF, the angle BKC is equal*" to the angle ELF : And because the circles ABC, DEF, are equal, the straight lines from their centres are equal : Therefore BK, KC are equal to EL, LF, and they contain equal angles : Therefore the base BC, is equal"^ to the base EF. There- fore, in equal circles, &c. Q. E. D. PROP. XXX. PROB. X O bisect a given circumference, that is, to du vide it into 4wo equal parts. Let ADB be the given circumference ; it is required to sect it. * 10. 1. Join AB, and bisect^ it in C ; from the point C draw at right angles to AB, and join AD, DB : The circumfa- rence ADB is bisected in the point D. Because AC is equal toCB, and CD common Ui the triangles ACD, BCD, the two sides AC, CD are equal to the two BC, CD ; and the angle ACD is equal to the angle BCD, because each of them is a right angle : Therefore the base AD is equal "4. 1. ^ to the base BD. But equal straight « 28. 3. lines cut off equal<^ circumferences, the greater equal to the greater, and the less to the less, and AD, DB are each of them less than a semicircle ; because DC passes through the cen- < Ccr. 1. 3. ^^^^ ' Wherefore the circumference AD is equal to tl?e cir- cumference DB : Therefore the given circumference is bi- sected in D, Which was to be done. 11 OF EUCLID. 85 Book III. PROP. XXXI. THEOR. '^^'^ 1 X a circle, the angle in a semicircle is a right an- gle ; but the angle in a segment greater than a se- micircle is less than a right angle ; and the angle in ' a segment less than a semicircle is greater than a right angle. Let ABCD be a circle, of which the diameter is BC, and centre E ; and draw CA, dividing the circle into the segments ABC, ADC, and join BA, AD, DC ; the angle in the semi- circle BAC is a right angle; and the angle in the segment ABC; which is greater than a semicircle, is less than a right angle ; and the angle in the segment ADC, which is less than a semicircle, is greater than a right angle. Join AE, and produce BA to F : and because BE is equal to EA, the arigle EAB is equal^ to EBA ; also, because AE '"'■ '" is equal to EC ; the angle EAC is equal to ECA ; wherefore ths whole angle BAC is equal to the two angles ABC, ACB: But FAC, the exterior angle of the triangle ABC, is equal'' to the two angles ABC, ACB : therefore the angle BAC is equal to the angle FAC, and each of them is therefore a righf^ angle : Wherefore the angle BAC in a semicircle is a right an- ^^ >/ ^lo. dcf. 1. g!e. And because the two angles ABC? BAC of the triangle ABC are together less"^ than two right angles, and that BAC s a right angle, ABC must be less than a right angle ; and :herefore the angle in a segment ABC greater than a semicir- :le, is less than a right angle. And because ABCD is a quadrilateral figure in a circle, any wo of its opposite angles are equal'' to two right angles :- < :herefore ths angles ABC, ADC are equal to two right an-* ^es ; and ABC is less than a right angle ; wherefore the other kDC is greater than a right angle. Besides, it is manifest, that the circumference of the greater segment ABC falls without the right angle CAB ; but the arcumference of tb€ less segment ADC faUs within the right gle CAP.' ' And this is all that is meant, when in the G 3 « Greek 36 Book III. THE ELEMENTS ' Greek text; and the translations from it, the angle of the ' greater segment is said to be greater, and the angle of the less ' segment is said to be less, than a right angle.' Cor. From this it is manifest, that if one angle of a triangle be equal to the other tv;o, it is a right angle, because the angle adjacent to it is equal to the same two; and when the adjacent angles ace equal, they are right angles. PROP. XXXII. THEOR. » :;. \. " 1?. 3. ■=31. 3. *32. 1. 'i::. 3. F a straight line touches a circle, and from the point of contact a straight line he drawn cutting the circle, the angles made by this line with the line touching the circle, shall be equal to the ^angles which are in the alternate segments of the circle. Let the straight line EF touch the circle ABCD in B, and from the point B let the straight Ijne BD be drawn, cutting the circle: The angles which BD makes with the touching line EF shall be equal to the angles in the alternate segmenis of the circle : that is, the angle FBD is equal to the angle which is in the. segment DAB, and the angle DBE to the aa- G^le in the se,^ment BCD. From the point B draw^' BA at right angles to EF, and taiaj any point C in the circumference BD, and join AD, DC, CB and because the straight line EF touchts the circle ABCDi; the point B, and BA is drawn at right angles to the touching line from the point of contact B, the centre of the circle is^ in BA ; therefore the angle ADB in a semi- circle is a righf^ angle and con<^e- quently theother two angles BAD, ABD are equal'^ to a right angle : But ABF is likewise a right angle : therefore the angle ABF is equal to the angles BAD, ABD: Pake from these equals the common angle ABD ; therefore the remaining dngle DBF is equal to the an- gle BAD, which is in the alternate segment of tlje circle j and because ABCD is a quadrilateral figure in a circle, the opposit^ angles BAD, BCD are equal*" to two right angles : therefo ths OF EUCLID. «7 the angles DBF, DBE, being likewise equaF to two right an- ^^^ in. gles, are equal to the angles BAD, BCD : and DBF has been f**^7*^ proved equal to BAD : Therefore the remaining angle DBE is equal to the angle BCD in the alternate segment of the cir- cle. Wherefore, if a straight line, &c. Q. £. D. PROP. XXXIII. PROB. U POX a ariven straio-btline to describe a seo-ment ot a circle, containing an angle equal to a given ^'=*^^'- rectilineal angle. Let AB be the given straight line, and the angle at C the given rectilineal angle ; it is required to describe upon the given straight line AB a segment of a circle, containing an angle equal to the angle C. First, let the angle at C be a right angle, and bisect* AB in F, aad from the centre F, at the distance FB, describe the semicircle AHB ; therefore the angle AHB in a semi- 10. 1. «" 31. 3. Ml.l. circle is*" equal to the right angle at C. But, if the angle C be not a right angle, at the point A, in the straight line AB, make"^ the angle BAD equal to the angle '23. i. C, and from the point A draw'^ AE at right angles to ^^ ADi bisect^ AB in F, and from F draw^ FG at right angles to AB, and join GB : And because AF is equal to FB, and FG common to the triangles AFG, BFG, the two sides AF, FG are equal to the two BF, FG ; and the angle AFG is equal to the angle BFG; therefore the base AG is equal^ to the base GB : and the circle described "^ **• from the centre G, at the distance GA, shall pass through the point B ; let this be the circle AHB : And because from the point A the extremity of the diameter AE AD is drawn at G 4 right 88 THE ELEMENTS ^0 oiclii. right angles to AE, therefore AD*^ touches the circle j and be- 'Cor. J6.5. cause AB drawn from the point of contact A cuts the circle, the angle DAB is equal to the angle «32, 3. in the alternate segment AriBs: But the angle D.\B is equal to the angle C, therefore also the angle C is equal to th.^ angle in the segment AHB : Wherefore, upon the given straight line AB the segment AHB of a circle is described, which contains an angle equal to the given angle at C. Which was to be done. PROP. XXXIV. PROB. » 17. 3. * 23. 1. e 32. 3. T O cut off a segment from a given shall contain an angle equal to a given circle which _ ^ ^^ rectilineal angle. Let ABC be the given circle, and D the given rectilineal angle ; it is required to cut off a segment from the circle ABC that shall contain an angle equal to the given angle D. Draw* the straight line EF touching the circle ABC in the point B, and at the p^int B, in the straight line BF, make'' the angle FBC equal to the angle D : Therefore, because the straight line EF touches the circle ABC, and BC is drawn from the point of contact B, the angle FBC is equal"^ to the angle in the alternate segment BAC of the circle; But the angle FBC is equal to the angle D:. therefore the angle in the segment BAC is equal to the angle D : Wherefore the segment BAC is cut off from the given circle ABC, containing an angle equal to the given angle D ; Which was to be done. OF EUCLID. PROP. XXXV. THEOR. 89 Uoox III. XF two straight lines within a circle cut one ano- ther, the rectangle contained hy the segments of one of them, is equal to the rectangle contained bj the segments of the other. Let the two straight lines AC, BD, within the circle ABCD, cut one another in the point E : the rechingle con- tained by AE, EC is equal to the redlangle contained by BE, ED. If AC, BD pass each of them through the centre, so that E is the centre ; it is evident, that AE, EC, BE, ED, being all equal, the redtangle AE, EC is likewise equal to the redangle BE, ED. But let one of them BD pass through the centre, and cut thfc other AC which does not pass through the centre, at right an- gles, in the point E : Then, if BD be bisected in F, F is the centre of the circle ABCD ; join AF : And because BD, which passes through the centre, cuts thcstraight line AC, which does not pass through the centre, at right angles in E, AE, EC are equal* to one another : And because the straight line BD is cut into two equal parts in the point F, and into two unequal in the point E, the rectangle BE, ED, together with the square of EF, is equal" to the square of FB ; that is, to the square of FA } but the squares of AE, EF axe equal ^ to the square of Fa ; therefore the reftangle BE, ED, together with the square of EF, is equal to the squares of AE, EF : Take away the common squareofEF, and the remaining rectangle BE, ED is equal to the remaining square of AE; that is, to the rectangle AE, EC, Next, let BD, which passes through the centre, cut the other AC, which does not pass through the centre, inE, but not at right angles : Then, as before, if BDbe bisected in F, F is fhe centre of the circle. Joio AF, and from F draw** FG per- pendicular Se«sr. »5.S. ••5.1. 47.1, * It. 1. fo THE ELEMENTS «47. 1. Book HI. pendicularto AC ; therefore AG is equal* to GC j wherefore VjT^s/ the re£bngle AE, EC, together with the square of EG is *5. 2. cquaP to the square of AG: To each of these equals add the square of GF j therefore the re£Vangle AE, EC, together with the squares of EG, GF is equal to the squares of AG, GF: But the squares of EG, GF are equal'^ to the square of EF ; "and the squares of AG, GF are e,qual to the square of AF : Therefore the rectangle AE, EC, together with the square of EF, is equal to the square of AF ; that is, to the square of FB : But the square of FB is equal^ to the reilangle BE, ED, together with the square of EF ; therefore the rectangle AE, EC, together with the square of EF, is equal to the redangle BE, ED, toge- ther with the square of EF : Take away the common square of EF, and the remaining redangle AE, EC, is therefore equal to the remaining rectangle BE, ED. ' Lastly, Let neither of the straight lines AC, BD pass through the centre ; Take the centre Sj and through E, the intersection of the straight lines AC, DB, draw the diameter GEFH : And because the redlangle AE, EC is equal, as has been shewn, 'to the reftangle GE, EH ; and, for the same reason, the rectangle BE, ED is equal to the same re£tangle GE, EH; therefore the re£tangle AE, EC is equal to the redangle BE, ED. Q,E. D. Wherefore, if two straight lines, &c. I PROP. XXXVl. THEOR. F from any point without a circle two straight lines be drawn, one of which cuts the circle, and the other touches it ; the rectangle contained by the whole line which cuts the circle, and the part of it without the circle, shall be equal to the square of the line which touches it. Let D be any point without the circle ABC, and DCA, DB two straight lines drawn from it,of which DCA cuts the circle, 2 an4 OF EUCLID. 91 18. 3. *>6.2. and DB touches the same : The reaangle AD, DC is equal ^^^^^ to the square of DB. Either DC A passes through the centre, or it does not ; first, let it pass through the centre E, and join EB j therefore the angle EBD is a right* angle : And be- cause the straight line AC is bise<5ied in E, and produced to the point D, the rectangle AD, DC, together with the square of EC, is equaP to the square of ED, and CE is ^qual to EB : Tbere- fore the reaangle AD, DC, together ij with the square of EB, is" equal to the square of ED: But the square of ED is equal'' to the squares of EB, BD, be- y / = 47. 1. cause EBD is a right angle : Therefore the reaangle AD^ DC, together with the square of £B, is equal to the squares of EB, BD: Take away the common square of EB ; therefore the remaining reftanglc AD, DC, is equal to the square of the tangent DB. But if DC A does not pass through the centre of the circle ABC, take "^ the centre E, and draw EF perpendicular^ to AC, anJjoinEB,EC,ED: And because the straight lineEF, which passes through the centre, cuts the straight line AC, which does not pass through the centre, at rigiit angles, it shall likewise bised^ y '3.3, it ; therefore AF is equal to FC : And because the straight line AC is bisefted in F, and produced to D, the reaangle AD, DC, together with the square of FC, is equal '' to the square of FD : To each of these equals add the square of FE ; therefore the reaangle AD, DC, together with the squares of CF, FE, is equal to the squares of DF, FE : But the square of ED is equal= to the squares of DF, FE, because EFD is a right an- gle : and the square of EC is equal to the squares of CF, FE ; therefore the reaangle AD, DC, to- gether with the square of EC, is equal to the square of ED: And CE is equal to EB ; therefore the reaangle AD, DC, to- gether with the square of EB, is equa.1 to the square of ED : But ° 1.3. «r2. J. 92 THE ELEMENTS ^^°^_^^2!J' ^"' '^^ squares of EB, BD are equal to the square"^ of ED, ^"^^yi""^ because EBD is a right angle ; therefore the rectangles AD, DC, together with the square of EB, is equal to the squares of EB, BD ; Take away the common square of EB ; there- fore the remaining redtangle AD, DC is equal to the square of DB. Wherefore, if from any point, &c. Q. E. D. Cor. If from any point without a circle, there be drawn two straight lines cutting it, as AB, AC, the rectangles contained by the whole lines and the parts of them without the circle, are equal to one another, viz. the redlan- 2;le BA, AE, to the rectangle CA, AF : For each of them is equal to the square ef the straight line AD which touches the circle. SeeN. « 17. 3. ^ IB. 3. <• 36. 3. PROP. XXXVII. THEOR. XF from a point without a circle there be drawn two straight lines, one of which cuts the circle, and the other meets it; if the rectangle contained by the whole line which cuts the circle, and the part of it without the circle be equal to the square of the line >vhich meets it, the line which meets shall touch the circle. Let any point D be taken without the circle ABC, and from it let two straight lines DCAand DB be drawn, of which DCA cuts the circle, and DB meets it; if the reClangle AD, DC be equal to the square of DB ; DB touches the circle. Draw* the sifaight line DE, touching the circle ABC, find its centre F, and join FE, FB, FD ; then ^ ED is a right'' an- gle : And beeause DE touches the circle ABC, and DCA cuts it, the recStangle AD, DC is equa^ to the square of DE : But the rectangle AD, DC is, by hypothesis, equal to the square of DB : Therefore the square of DE is equal to the square of DB j and the straight line DE equal to the straight line DB : And fe; OF EUCLID. 93 !8. 1, FE is equal to FB, wherefore DE, EF are equal to DB, BF; ^^?^* and the base FD is common to the '' two triangles DEF, DBF ; therefore the angle DEF is equal** to the angle DBF; but DEF is a right angle, therefore also DBF is a right angle : And FB, if produced, is a diameter, and the straight line which is drawn at right angles to a diameter, from the extremity of it, touches' the cir- cle : Therefore DB touches the circle ABC. Wherefore, if from a point, &c. Q. E.-D. 16>3. [ 94- ] THE ELEMENTS EUCLID. BOOK. IV. DEFINITIONS. I. r^^,^ jLx. Rectilineal figure is said to be inscribed in another SteNi redlilineal figure, when all the angles of the inscribed fi- gure are upon the sides of the figure in which it is inscribed, each upon each; II. In like manner, a figure is said to be described about another figure, when ail the sides of the circumscribed figure pass through the angular points of the figure about which it is described, each through each. III. A re£lilineal figure is said to be inscribed in a circle, when all the angles of the inscribed figure are upon the circumfe- rence of the circle. IV. A rectilineal figure is said to be described about when each side of the circumscribed figure touches the circumference of the circle. V. In like manner, a circle is said to be in- scribed in a redlilineal figure, when the circumference of the circle touches each side of the fiiiurc. THE ELEMENTS OF EUCLID VL A circle is said to be described about a rec- tilineal figure, when the circumference of the circle passes through all the angu- .lar points of the figure about which it is described. VIL A straight line is said to be placed in a cir- cle, when the extremities of it are in the circumference of the circle. PROP. L PROS. In a given circle to place a straight line, equal to a given straight line not greater "than the diameter of the circle. Let ABC be the given circle, and D the given straight line, not greater than the diameter of the circle. Draw BC the diameter of the circle ABC j then, if BC is equal to D, the thing required is done ; for in the circle ABC a straight line BC is placed equal to D : But, if it is not, BC is greater than D j make CE equal* to D, and from the centre C, at the distance CE, describe the circle AEF, and join CA: Therefore, because C is the centre of the circle AEF, CA is equal to CE ; but D is equal to CE ; therefore D is equal to CA: Where- fore in the circle ABC, a straight line is placed equal to the given straight line D, which is not greater than the diameter •f the wrcle. Which was to be done. PROP. U. PROB. IX'a given circle to inscribe a triangle equiangu- lar to a given triangle. i-et » 17. 3. *23. J. « 32. 3. '»32. 1. THE ELEMENTS Let ABC be the given circle, and DEF the given trian- gle ; it is required to inscribe in the circle ABC a triangle equiangular to the triangle DEF. Dravi'^ the straight line GAH touching the circle in the point A, and at the point A, in the straight line AH, make" the angle HAC equal to the angle DEF ; and at the point A in the straight line AG, n?ake the angle GAB equal to the angle DFE, and join BC : There- ibre, because HAG touches the circle ABC, and AC is drawn from the point of contact, the angle HAC is equal'^ to the angle ABC in the alternate segment of the circle: But HAC is equal to the an- gle DEF : therefore also the angle ABC is equal to DEF : For the same reason, the angle ACB is equal to the angle DFE ; therefore the remaining angle BAC is equaF to the remaining angle EDF : Wherefore the triangle ABC is equiangular to the triangle DEF, and it is inscribed in the circle ABC. Which was to be done. PROP. HI. PROB. »«3. I. 17.3. <^ 18, 3. About a given circle to describe equiangular to a given triangle. triangle Let ABC be the given circle, and DEF the given triangle; it is required to describe a triangle about the circle ABC equiangular to the triangle DEF. Produce EF both ways to the points, G, H, and jRnd the centre K of the circle ABC, and from it draw any straight line KB ; at the point K, in the straight line KB, make* the angle BKA equal to the angle DEG, and the angle BKC equal to the angle DP'H ; and through the points A, B, C, draw the straight lines LAM,"MBN, NCL, touching'' the circle ABC : Therefore, because LM, MN, NL touch the circle ABC in the points A, B, C, to which from the centre are drawn KA, KB, KC, the angles at the points A, B, G, are righf^ angles : And because the four angles of the quadrilateral figure AMBK are equal OF EUCLID. 97 Kqual to four right angles, for it can be divided into two trian- Book IY. gles ; and that two of them KAM, KBM are right angles, the '^•^''^^''*^^ other two AK.B, AMB are equal to two right angles : But the angles DEG, DEF are likewise equal" to two right angles ; therefore the an- gles AKB, AMB are equal to the angles DEG, DEF, of which AKB is equal to DEG ; wherefore the remaining angle'AMB is e'qual to the remaining angle DEF : In like manner, the an- gle LNM may be demonstrated to be equal to DFE ; and therefore the remaining angle MEN is equal'' to the remain- e 22. i, ing angle EDF: Wherefore the triangle LMN is equiangu- lar to the triangle DEF : And it is described about the circle ABC. Which was to be done. PROP. IV. PROB. O inscribe a circle in a siven trian ole. Se«N. Let the given triangle be ABC j it is required to inscribe a circle in ABC, Bised* the angles ABC, BCA by the straight lines BD, CD ^ ^ ^ meeting one another in the point D, from which draw** DE, ^ ^3. 1. DF, DG perpendiculars to AB, BC, CA : And because the angle EBD is equal to the angle FBD, for the angle ABC is biseded by BD, and that the right angle BED is equal to the right angle BFD, the two triangles EB'D, FBD have two angles of the one equal to two angles of the other, ^and the side BD, v/hich is oppo- site to one of the equal angles in each, is common to both ; there- fore their ether sides shall b» H equal ; ^8 Book IV. 26. 1. "16.3. THE ELEMENTS equa^ } wherefore DE is equal to DF : For the same reason, DG is equal to DF ; therefore the three straight linesDE, DF, DG are equal to one another, and the circle described from the centre D, at the distance of any of them, shall pass through the extremities of the other two, and touch the straight lines AB, BC, CA, because the angles at the points E, F, G, are right angles, and the straight line which is drawn from the extremity of a diameter at right angles to.it, touches'* the cir- cle: Therefore the straight lines AB, BC, CA do each of them touch the circle, and the circle EFG is inscribed in the triane;le ABC. Which was to be done. PROP. V. PROB. SecN ' j[ o describe a circle about a given triangle. Let the given triangle be ABC j it is required to describe a circle about ABC. , )o_ 1 Biseft* AB, AC in the points D, E, and from these points » 11. u draw DP^,EF at right angles'' to AB, AC ; DF, EF produced C B «*. 1. meet one another : For, if they do not meet, they are parallel wherefore AB, AC, which are at right angles to them, are pa rallel ; which is absurd : Let them meet in F, and join FA also if the point F be not in BC, join BF, CF : Then, bccaus( . AD is equal to DB, and DF common, and at right angles t( AB, the base AF is equal*= to the base FB . In like manner, i may be shown that CF is equal to FA ; and therefore BF f equal to FC i and J'A, FB, FC are equal to one anothef wherefor OF EUCLID. 99 wherefore the circle described from the centre F, at the dis- ^"o"^ ^• tance of one of them, shall pass through the extremities of the ^"^ other two, and be described about the triangle ABC. Which was to be done. Cor. And it is manifest, that when the centre of the circle falls within the triangle, each of its angles is less than a right angle, each of them being in a segment greater than a semicir- cle j but, when the centre is in one of the sides of the trian- gle, the angle opposite to this side, being in a semicircle, is a right angle ; and, if the centre falls without the triangle, the angle opposite to the side beyond which it is, being in a seg- ment less than a semicircle, is greater than a right angle : Wherefore, if the given triangle be acute angled, the centre of the circle falls within it ; if it be a right angled triangle, the centre is in the side opposite to the right a- .gle ; and, if it be an obtuse angled triangle, the centre fails without the tri- angle, beyond the side opposite to the obtuse angle. PROP. VI. PROB. 1 O inscribe a square in a given circle. Let ABCD be the given circle j it is required to inscribe a square in ABCD. Draw the diameters AC, BD, at right angles to one another; .and join AB, BC, CD, DA; because BE is equal to ED, for E is the centre, and that EA-is com- mon, and at right angles to BD ; the base BA is equal* to the base AD ; and, for the same reason, BC, CD are each of them equal to BA, or AD ; therefore the quadrilateral fi- gure ABCD is equilateral. It is al- o rectangular ; for the straight line ;D, being the diameter of the circle iiBCD, BAD is a semicircle ; where- fore the angle BAD is a right'' an- te i for the same reason each of the angles ABC, BCD, CDA, a right angle; therefore the quadrilateral figure ABCi^ is rectangular, and it has been shown to be equilateral; there- fore it IS a square ; and it is inscribed in the circle ABCD ' v'hich was to be done. 4. I, '31. I H2 • Its. 3. '> 1». 3. e.28.1. « 54. 1. THE ELEMENTS PROP. VII. PROB. X O describe a square about a given circle. Let ABCD be the given circle j it is required to describe a square about it. Draw two diameters AC, BD of the circle ABCD, at right angles to one another, and through the points A, B, C, D, draw* FG, GH, HK, KF touching the circle ; and because FG touches the circle ABCD, and EA is drawn from the centre E to the point of contadl A, the angles at A are right'' angles ; for the same reason, the angles at the points B, C, D are right angles ; and because the angle AER is a right angle, as liicewise is EBG, GH is parallel"^ to AC ; for the same rea- son, AC is parallel to FK, and in like manner GF, HK may each of them be demonstrated to be parallel to BED ; therefore the figures GK, GC, AK, FB, BK are parallelograms ; andGF is therefore equaH to HK, and GH to ¥K ; and because AC is equal to BD, and that AC is equal to each of the two GH, FK ; and BD to each of the two GF, HK ; GH, FK are each of them equal to GF or HK ; therefore the quadrilateral figure FGHK i> equilateral. It is also rectangular ; for GBEA being a parallel- ogram, and AEB a right angle, AGS'* is liicewise a right an- gle: In the same manner it may be shown that the angles at H, K, F are right angles j therefore the quadrilateral figure FGHK is rectangular, and it was demonstrated to be equila- teral ; therefore it is a square ; and it is described about the ckcle ABCD, Which was to be done. B a A F E 1 s y a c K »10. 1. ••31.1. PROP. Vin. PROB. X O inscribe a circle in a given square. Let ABCD be the given square ; it is required to inscribe a circle in ABCD. Bisect* each of the sides AB, AD, in the points F, E, and through E draw ^ EH parallel to AB or DC, and through F drav( I OF EUCLID. 101 draw FK parallel to AD or BC; therefore each of the figures AK, BookIV. KB, AH, HD, AG, GC, BG, GD, is a parallelogram, and ""p"^*^ their opposite sides are equal''; and because AD is equal to AB, ^ "^■*' ^• and that AE is the half of AD, and' AF the half of AB, AE is equal to AF ; wherefore the sides opposite to these are equal, viz. FG to A E D GE ; in the same manner it may be de- monstrated that GH, GK. are each of them equal to FG or GE : therefore the four straight lines GE, GF, GH, F| ^ JK. GK are equal to one another; and the circle described from the centre G at the distance of one of them, shall pass _ _ through the extremities of the other B H three, and touch the straight lines AB, BC, CD, DA; because the angles at the points E, F, H, K, are right "^ angles, and that the scraight line which is drawn from "29.1. the extremity of a diameter, at right angles to it, touches the circle*^; therefore each of the straight lines, AB, BC, CD, '^^•^• DA touches the circle, which therefore is inscribed in the square ABCD. Which was to be done. PROP. IX. PROB. X O describe a circle about a given square. Let ABCD be the given square j it is required to describe a circle about it. Join AC, BD, cutting one another in E ; and because DA is •qual to AB, and AC common to the triangles DAC, BAG, the two sides DA, AC are equal to the two BA, AC, and the base DC is equal to the base BC ; wherefore the angle DAC is equal^ to the angle BAC, and the angle DAB is bisected by the straight line AC : In the same manner, it may be demonstrated that the angles ABC, BCD, CDA are severally bisected bythe straight lines BD, AC ; therefore, because the angle DAB is equal to the angle ABC, and that the an^le EAB is the half of DAB, and EBA the half of ABC: The angle EAB is equal to the angle EBA ; wherefore the side JEA is equal'' to the side EB : In the same manner, it may be *^' ^* H 3 d«monstratait. 8.1. io2 THE ELEMENTS Book IV. demonstrated, that the straight lines EC, ED are each of ^'*^^''*^ them equal to EA or EB ; therefore the four straight lines EA, EB, EC, ED, are equal to one another ; and the circle described from the centre E, at the distance of one of them, shall pass through the extremities of the other three, and be described about the square ABCD, Which was to be done. »11. 2. *1. 4. <: 5. 4. « 37. 3. • »«. 3. ''32.1. PROP. X. PROB. A O describe an isosceles triangle, having each of the angles at the base double of the third angle. Take any straight line AB, and divide* it in the point C, so that the redlangle AB, BC be equal to the square of CA ; and from the centre A, at the distance AB, describe the circle BDE, in which place'' the straight line BD equal to AC, which is not greater than the diameter of the circle BDE j join DA, DC, and about the triangle ADC describe^ the circle ACD j the triangle ABD is such as is required, that is, each of the angles ABD, ADB is double of the angle BAD. . . . Because the re6langle AB, BC is equal to the square of AC, and that AC is equal to BD, the reilangle AB, BC is equal to the square of BD ; and because from the point B, without the circle ACD, two straight lines BCA, BD are drawn to the cir- cumference, one of which cuts, and the other meets the circle, and that the rectangle AB, BC, contained by the whole of the cutting line, and the part of it without the circle, is equal to the square of BD which meets it ; the straight line BD touches^ the circle ACD ; and because BD touches the circle, and DC is drawn from the point of con- ta€t D, the angle BDOis equal«= to the angle DAC in the alternate segment of the circle ; to each of these add the angle CDA ; therefore the whole angle BDA is equal to the two angles CDA, DAC ; but the exterior angle BCD is equaF to the angle CDA, DAC j therefore also BDA is equal to BCD ; but OF EUCLID. 103 but BDA is equal? to the angle CBD,because the side AD ^°°^ ^^' is equal to the side AB ; therefore CBD, or DBA", is equal to g^"^Y^ BCD ; and consequently the three angles BDA, DBA, BCD, are equal to one another ; and because the angle DBC is equal to the angle BCD, the side BD is equaP to the side DCj but ''^- ^• BD was made equal to CA, therefore also CA is equal to CD, and the angle CDA-equal? to the angle D AC ; there- fore the angles CDA, DAC together, are double of the angle DAC : But BCD is equal to the angles CDA, DAC ; there- fore also BCD is double of DAC, and BCD is equal to each of the angles BDA, DBA j each therefore of the angles BDA, DBA is double of the angle DAB wherefore an isosceles tri- angle ABD is described, having each of the angles at the base double of the third an^le. Which was to be done. PROP XI. PROB. X O inscribe an equilateral and et|uiangular pen- tagon in a given circle. Let ABCDE be the given circle; it is required to inscribe an equilateral and equiangular pentagon in the circle ABCDE. Describe^an isosceles triangle FGH, having each of the an- * ^^- *• gles at G, H, double ot the angle at F j and in the circle ABCDE inscribe'' the triangle ACD equiangular to the trian- '' 2. *• gle FGH, so that the angle CAD be equal to the angle j^ at F, and each of the angles ACD, CDA equal to the angle at G or H, wherefore each of the angles ACD, CDA is double of the angle CAD. Bisect'^ the angles ACD, CDA by the straight lines CE, DB j and join AB, BC, DE, EA. ABCDE is the pentagon required. Because each of the angles ACD, CDA is double of CAD, and are bisected by the straight lines CE, DB, the'five angles DAC, ACE, ECD, CDB, BDA are equal to one another ; but equal angles stand upon equal"* circumferences ;' therefore « 26. ». the five circumferences A B, BC, CD, DE, EA are equal to one H4 another: 9. 1. 104 * THE ELEMENTS Book IV. another : And equal circumferences are subtended by equal' ^^^3^ Straight lines J therefore the five straight lines AB, BC, CD, DE, EA are equal to one another. Wherefore the pentagon ABCDE is equilateral. It is also equiangular j because the circumference AB is equal to the circumference DE: If to- each be added BCD, the whole A BCD is equal to the whole EDCB : And the angle AED stands on the circumference ABCD, and the angle BAE on the circumference EDCB; f2^, 3. therefore the angle BAE is equaF to the angle AED : For the same reason, each of the angles ABC, BCD, CDE is equal to the angle BAE, or AED: Therefore the pentagon ABCDE is equiangular ; and it has been shown that it is equilateral. Wheretore, in the given circle, an equilateral and equiangular pentagon has been inscribed. Which was to be done. PROP. XII. PROB. X O tlescribe an equilateral and equiangular pen- tagon about a given circle. Let ABCDE be the given circle ; it is required to describe an equilateral and equiangular pentagon about the circle ABCDE. Let the angles of a pentagon, inscribed in the circle, by the last proposition, be in the points A, B, C, D, E,sothat thecir- •^ n. 4. cumferences AB, BC, CD, DE, EA are equal'' ; and throujjh the points A, B, C, D, E, draw GH, HK, KL, LM, Ut}, * 17. 3. touching'' the circle ; take the centre F, and join FB, FK, FC, FL, FD : And because the straight line KL touches the circle ABCDE in the point C, to which FC is drawn from the ccn- f 18. 3. tre F, FC is perpendicular"" to KL ; therefore each of the an- gles at C is a right angle : For the same reason, the angles at the points B, D are right angles : And because FCK is a right "47. 1. angle, the square of FK is equal"" to the squares of FC, CK ; For the same reason, the square of FK is equal to the squares of FB, BK : Therefore the squares of FC, CK are equal to the squares of FB, BK, of which the square of FC is equal to the square gf FB ; the remaining squar* of CK is therefore equal to the OF EUCLID. 105 the remaining square of BK, and the straight line CK equal to BookIV, BK : And because FB is equal to FC, and FK common to the ^"^^''^'^ triangles BFK,CFK,thetwoBF,FKare equal tothetwoCF, FK j'and the base BK is equal to the base KC ; therefore the angle BFK is equal= to the angle KFC, and the angle BKF to -^S. I. FKC i wherefore the angle BFC is double of the angle KFC, and BK.C double of FKC": For the same reason, the angle CFD is double of the angle CFL, and CLD double of CLF : And be- cause the circumference BC is equal to the circumference CD, the angle BFC is equaF to the ^ f 27. 3. angle CFD ; and BFC is dou- ble of the an^le KFC, and CFD double ot CFL ; there- fore the angle Kr C, is equal to the angle CFL : and the right angle FCK is equal to the right angle FCL : Therefore, in the twotrianglesFKCjFLC, there are two angles of one equal to two angles of the other, each toeach, and the side FC, which is adjacent to the equal angles in each, is common to both; therefore the other sides shall be equals to the other sides, and ^26. 1. the third angle to the third angle : fherefore the straight line KC is equal to CL, and the angle FKC to the angle FLC : And because KC is equal to CL, KL is double of KC : In the same manner it may be shown that HK is double of BK : And because BK is equal to KC, as was demonstrated, and that KL is double of KC, and HK double of BK, HK shall be equal to KL : In like manner, it may be shown that GH, GM, ML are each of them equal to HI^ orKL: Therefore the pentagon GHKLM is equilateral. It is also equiangular ; for, since the angle FKC is equal to the angle FLC, and that the angle UKLis double oftheangleFKC, and KLMdoubleof FLC, as was before demonstrated, the angle HKL is equal to KLM: And in like manner it may be shown, that each of the angles KHG, HGM, GML is equal to the angle HKL or KLM : Therefore the five angles GHK, HKL, KLM, LMG MGH being equal to one another, the pentagon GHKLM is equian- gular : And it is equilateral, as was demonstrated j and it is de« jcribed about the circle ABCDE. Which was to be done. •9.1. *4. 1. •!K.l. THE ELEMENTS PROP. XIII. PROB. 1 O inscribe a circle in a given equilateral and equiangular pentagon. Let ABCDE be the given equilateral and equiangular penta- gon ; it is required to inscribe a circle in the pentagon ABCDE. Bisect* the angles BCD, CDE by the straight lines CF DF, and from the point F, in which they meet, drav/ the straight lines FB, FA, FE : Therefore since BC is equal to CD, and CF common to the triangles BCF, DCF the two sides BC, CF are equal to the two DC,CF ; and the angle BCF is equal to the angle DCF ; therefore the base BF is equal'' to the base FD, and the other angles to the other angles, to which the equal sides are opposite ; therefore the angle CBF is equal to the angle CDF : And because the angle CDE is double of CDF, and that CDE is equal to CBA, and CDF to CBF; CBA is also double of the angle CBF; there- fore the angle ABF is equal to the angle CBF ; wherefore the angle ABC is bisected by the straight line BF ; In the same manner it may be demonstrated, that the angles BAE, AED, are bisected by the straight lines AF, FE: From the point F draw^ FG, FH, FK, FL, FM perpendiculars to the straight lines AB, BC, CD, DE, EA : And because the angle HCF is equal to KCP, and the right angle FHC equal to the right angle FKC ; in the triangles FHC, FKC there are two angles of one equal to two angles of the other, and the side FC, which is opposite to one of the equal angles in each, is common to both ; therefore the other sides shall be equal^, each to each; wherefore the per- pendicular FH is equal to the perpendicular FK : In the same manner it may be demonstrated ; that FL, FM, FG are each of them equal to FH or FK : Therefore the five straight lines FG, FH, FK, FL, FM are equal to one another : ' Wherefore the circle described from the centre F, at the dis- tance of one of these five, shall pass through the extremities of 2 the.' OF EUCLID. 107 the other four, and touch the straight lines AB, BC, CD, DE, Book iv, EA, because the angles at the points G, H, K, L, M are ^"^^^^^ right angles ; and that a straight line drawn from the extre- mity of the diameter of a circle at right angles to it, touches^ * 16. 3, the circle : Therefor*each of the straight lines AB, BC, CD, DE, EA touches the Circle : wherefore it is inscribed in the pentagon ABCDE. Which was to be done. PROP. XIV. PROB. A O describe a circle about a given equilateral and equiangular pentagon. Let ABCDE be the given equilateral and equiangular pen- tagon : it is required to describe a circle about it. Bisea=' the angles BCD, CDE by the straight lines CF, FD, • 9. 1. and from the point F, in which they meet, draw the straight iines FB^ FA, FE to the points B, A, E. It may be demonstrated, in the same manner as in the preceding proposition, that the angles CBA, BAE, AED are biseded by the straight lines FB, FA, FE: And because the angle BCD is equal to . the angle CDE, and that FCD is the half of the angle BCD, and CDF, the half of CBE ; the angle FCD is equal to FDC ; wherefore the side CF is equaP to the side FD : In like manner it may be de- " 5. i monstrated that FB, FA, FE, are each of them equal to FC or FD : Therefore the five straight lines FA, FB, FC, FD, FE are equal to one another j and the circle described from the centre F, at the distance of one of them, shall pass through the extremities of the other four, and be described about the equilateral and equiangular pentagon ABCDE. Which was to be done. THE ELEMENTS PROP. XV. PROB. seeN. X O inscribe an equilateral and equiangular hexa- gon in a given circle. Let ABCDEF be the given circle j it is required to inscribe an equilateral and equiangular hexagon in it. Find the centre G of the circle ABCDEF, and draw the di- ameter AGD J and from D as a centre, at the distance DG, describe the circle FGCH, join EG, CG, and produce them to the points B, F ; and join AB, BC, CD, DE, EF, FA : The hexagon ABCDEF is equilateral and equiangular. Because G is the centreof the circle ABCDEF, GE is equal to GD : And because D is the centre of the circle EGCH, DE is equal to DG ; wherefore G£ is equal to ED, and the tri- angle FGD is equilateral; and therefore its three angles EGD, GDE, DEG, are equal to one another, because the angles at the base of an isosceles triangle are equaP; and the three angles of a triangle are equal'' to two right angles ; therefore th» angle EGD is the third part of two right angles; In the same manner it may be demonstrated, that the angle DGC is also the third part of two right angles : And because the straight line GC makes with EB the adjacent angles EGC, CGB equaP to two right angles; the remaining angle CGB is the third part of two right angles ; therefore the angles EGD, DGC, CGB are equal to one another : And to these are equal'' the vertical opposite angles BGA, AGF, FGE : I'herefore the six angles EGD, DGC, CGB, BGA, AGF, FGE, are equal to one another : But equal angles stand upon equal'= circumfe- rences ; therefore the six circumfe- rences AB, BC, CD, DE, EF, FA are equal to one another . And equal circumferences are subtended by equal*" straight lines ; therefore the six straight lines are equal to one another, and the hexagon ABCDEF is equilateral. It is also equiangu- lar : for, since the circumference AF is equal to ED, to each of | these add the circumference ABCD ; therefore the whole cir- cumference FABCD shall be equal to the whole EDCBA : An«i »5. 1. »32. i. "=13. 1. * 15. 1. 26.3. ^29.3. OF EUCLID 109 And the angle FED stands upon the circumference FABCD, Book rv. and the angle AF£ upon EDCBA ; therefore the angle AF£ ^■^''''*^. is equal to FED : In the same manner it mav he demonstrated that the other angles of the hexagon ABCDEF are each of them equal to the angle AFE or FED : Therefore the hexa« gon is equiangular; and it is equilateral, as was shown; and it is inscribed in the given circle ABCDEF. Which was to be done. Cor. From this it is manifest, that the side of the hexajon is equal to the straight line from the centre, that is, to the semidiameter of the circle. And if through the points A, B, C, D, E, F there be drawn straight lines touching the circle, an equilateral and equiangu- lar hexagon shall be described about it, which may be demon- strated from what has been said of the pentagon ; and likewise a circle may be inscribed in a given equilateral and equiangu- lar hexagon, and circumscribed about it, by a method like to that used for the pentagon. PROP. XVI. PROB. L O inscribe an equilateral and equiangular quin- seeN. decagon in a given circle. Let ABCD be the given circle ; it is required to inscribe an equilateral and equiangular quindecagon in the circle ABCD. Let AC be the side of an equilateral triangle inscribed* in ,2 4 the circle, and AB the side of an equilateral and equiangular pentagon inscribed'' in the same ; therefore, of such equal parts h u 4 as the whole circumference ABCDF contains fifteen, the cir- cumference ABC, being the third part of the whole, contains five; and the circumference AB, which is the fifth part of the whole, contains three ; therefore BC their difference contains tv/o of the same parts : Bi- aea^BCinE; therefore BE, EC ""W V/ ^30.3. are, each of them, the fifteenth part ©f the whole circumference ABCD: Therefore, if the straight lines BE, EC be drawn, and straight lines equal to them- be placed round'' m the whole circle, an equilateral and equiangular qum- d 1 4 <iecagon shall be inscribed in it. Which was to be done. And, no Book IV. THE ELEMENTS OF EUCLID. And, in the same manner as was done in the pentagon, if, through the points of division made by inscribing the quinde- cagon, straight lines be drawn touching the circle, an equila- teral and equiangular quindecagon shall be described about it; And likewise, as in the pentagon, a circle may be inscribed in a given equilateral and equiangular quindecagon, and circum- scribed about it. ' A [ iii ] THE E L E AI E X T S OF EUCLID. BOOK V. DEFINITIONS. I. LESS magnitude is said to be a part of a greater mag- Book V. nitude, ,^hen the less measures the greater y that is, v^»v-^/ 'when the less is contained a certain number of times ex- ' actly in the greater.' A greater magnitude is said to be a multiple of a less, when the greater is measured by the less ; that is, ' when the greater * contains the less a certain numl>er of times exactly.* III. * Ratio is a mutual relation of two magnitudes of the same See N. kind to one another, in respecl of quantity.' IV. Magnitudes are said to have a ratio to one another, when the less can be multiplied so as to exceed the other. V. The first of four magnitudes is said to have the same ratio to the second, which the third has to the fourth, when any equimultiples whatsoever ef the first and third being taken, and any equimultiples whatsoever of the second and fourth ; if the multiple of the first be less than that of the second, the multiple of the third is also less than that of the fourth ; or, if the multiple of the first be equal to that of the second, the multiple of the third is also equal to that of the fourth ; or. Hi THE ELEME^fTS. Book V. orj if the multiple of the first be greater than that of the se- '^■^'^■'^^^ cond, the multiple of the third is also greater than that of the fourth. VI. Magnitudes which have the same ratio are called proportion- als. N. B. ' When four magnitudes are proportionals, it is * usually exprest by saying, the first is to the second, as the * third to the fourth.' VII. When of the equimultiples of four magnitudes (taken as in the fifth definition), the multiple of the first is greater than that of the second, but the multiple of the third is not greater than the multiple of the fourth ; then the first is said to have to the second a greater ratio than the third magnitude has to the fourth ; and, on the contrary, . the third is said to have to the fourth a less ratio than the first has to the second. VIII. AnalogVj or proportion, is the similitude of ratios. Proportion consists in three terms at least. X. See ^t When three magnitudes are proportionals, the first is said to have to the third the duplicate ratio of that which it has to the second. XI. When four magnitudes are continual proportionals, the first is said to have to the fourth the triplicate ratio of that which it has to the second, and so on, qnadruplicate, See. increasing the denomination still by unity, in any number of propor- ' tionals. Definition A, to wit, of compound ratio. When there are any number of magnitudes of the same kind, the first is said to have to the last of them the ratio com- pounded of the ratio which the first has to the second, and of the ratio which the second has to the third, and pf the ratio which the third 'has to the fourth, and soon unto the last magnitude. For example, if A, B, C, D, be four magnitudes of the sam kind, the first A is said to have to the last D the ratio com-! pounded of the ratio of A to,B, and the ratio of B to C, and of the ratio C to D, or, the ratio of A to D is said to be compounded of the ratios of A to B, B to C, and C to D : And 1 ' OF EUCLID. n3 And if A has to B the same ratio which E has to F ; and I^ Book \ . to C, the same ratio that G has to H ; and C to D, the same '""^^•'^^ that K has to L ; then, by this definition, A is said to have to D the ratid compounded of ratios which are the same with the ratios of E to F, G to H, and K to L : And the same thing is to be understood when it is more briefly expressed, by saying, A has to D the ratio compounded of the ratios of E to F, G to H, and K to L. In like manner, the same things being supposed, if M has to N the same ratio which A has to D j then, tor shortness sake, M is said to have to N, the ratia compounded of the ratios of E to F, G to H, and K to L. XII. In proportionals, the antecedent terms are called homologous to one another, as also the consequents to one another. * Geometers make use of the following technical words to sig- ' nify certain ways of changing either the order or magni- ' tude of proportionals, so as that they continue still to be * proportionals.' XIII. Permutando, or alternando, by permutation, or alternately. This word is used when there are four proportionals, and it s« N. is inferred, that the first has the same ratio to the third, which the second has to the fourth ; or that the first is to the third, as the second to the fourth : As is shewn in the 1 6th prop, of this 5th book. XIV. Invertendo, by inversion : When there are four proportionals, and it is inferred, that the second is to the first, as the fourth to the third. Prop. B. Book 5. XV. .Componendo, by composition j when there are four propor- tionals, and it is inferred, that the first, together with the second, is to the second, as the third, together with the fourth, is to the fourth. i8th Prop. Book 5. XVI. Dividendo, by division j when there are four proportionals, and it is inferred, that the excess of the first a&ve the second, is to the second, as the excess of the third above the fourth, is to the fourth. 17th Prop. Book 5. XVII. Convertendo, by conversion ; when there are four proportion- als, and it is inferred, that the first is to its excess above the I second, 114 THE ELEMENTS SooK y. second, as the third to its excess above the fourth. Prop. "^-^^^^ E. Book 5. . • XVIIl. Ex oequali (sc. distantia)^ or ex sequo, from equality of dis- tance i when there is any number of magnitudes more than two, and as many others, so that they are proportionals when taken two and two of each rank, and it is inferred, that the first is to the last of the first rank of magnitudes, as the first is to the last of the others : * Of this there are the two fol- * lowing kinds, which arise from the different order in which * the magnitudes are taken, two and two.' XIX. ■ Ex oequali, from equality. This term is used simply by Itself, when the first magnitude is to the second of the first rank, as the first to the second of the other rank ; and as the second is to the third of the first rank, so is the second to the third of the other j and so on in order, and the inference is as men- tioned in the preceding definition ; whence this is called or- dinate proportion. It is demonstrated in 22nd Prop. Book 5. XX. Ex aequali, in proportione perturbata, seu inordinata, from equality, in perturbate or disorderly proportion*. This term is used when the first magnitude is to the second of the first rank, as the last but one is to the last of the second^ rank ; and ^s the second is to the third of the first rank, so is the last but two to the last but one of the second rank and as the third is to the fourth of the first rank, so is th< third from the last to the last but two of the Second rank -3 and so on in a cross order : And the inference is as in tl 1 8th definition. It is demonstrated in the 23d -Prop. Book 5. AXIOMS. I. JtLqyiMtJLTiPLES of the same, or of equal magnitudes, arwi equal to one another. ^f. II. Those' * 4 Prop, lib, 2. Archimedis dc tphira et cjlindro. OF EUCLID. 115 IJ, Book V. lose magnitudes of which the same, or equal magnitudes, '^^^"^ are equimultiples, are equal to one another. III. multiple of a greater magnitude is greater than the same multiple of a less. IV. hat magnitude of which a multiple is greater than the same multiple of another, is greater than that other magnitude. PROP. I. THEOR. G H D F any number of magnitudes be equimultiples of many, each of each ; what multiple soever any e of them is of its part, the same multiple shall all ; first magnitudes be of all the other. et any number of magnitudes AB, CD be equimultiples AS many others E, F, each of each ; whatsoever multiple is of E, the same multiple shall AB and CD together be and F together, lecause AB is the same multiple of E that CD is of F, as y magnitudes as are in AB equal to E, so many are there D equal to F. Divide AB into magnitudes il to E, viz. AG, GB ; and CD into CH, equal each of them to F : The number jfore of the magnitudes CH, HD shall be I to the number of the others AG, GB : because AG is equal to E, and CH to F, jfore AG and CH together are equal to» E B ' Ax. 2. 5. F together : F«r th^ same reason, because 's equal to E, and HD to Fj GB and HD her are equal to E and F together. Where- as many magnitudes as are in AB equal so many are there in AB, CD together to E and F together. Therefore, what- r multiple AB is of E, the same mul- is AB and CD together of E and F to- r. erefore, if any magnitudes, how many soever, be equi- les of as many, each of each, whatsoever multiple any ^ them is of its part, the same multiple shall all the first ' udes be of all the other: * For the same demonstration I 2 ' holds it6 THE ELEMENTS Book V. ( holds in anv number of maguuuaes, which was here applied ''•^•^^ t to two.' Q. E. D. PROP. II. THEOR. I If the fii-st magnitude be the same multiple of th( second that the third is of the fourth, and the fiftl the same multiple of the second that the sixth is o -the fourth ; then shall the first together with tb fifth be. the same multiple of the second, that th third together with the sixth is of the fourth. Let AB the first, be the same nt^ultiple of C the second, th; -DE the third is of F the fourth ; and BG the fifth, the san 3 multiple of C the second, that EH the sixth is of F the fourth : Then is AG the first, together with the fifth, the same multiple of C the second, that'DH the third, together with the sixth, is of F the fourth. Because AB is the same multiple of C, that DE is of F i there are as many magnitudes in AB equal to C, as there are in DE equal to F : In like manner, as many as there are in BG equal to C, so many there in EH equal to F : As many, then, as are in the wl AG equal to C, so many are there in the whole DH eqy^ F J therefore AG is the same multiple of C, that DH isp! that is, AG the first and fifth together, is the same multiple of the second C, that DH the third and sixth together is of the fourth F. If therefore, the first be the same multiple, &c. Q^ E. D. Dl E H A Cor. * From this it is plain, that, if any 'number of magnitudes AB, BG, GH, * be multiples of another C j and as many * DE, EK, KL be the same multiples of * F, each of each j the whole of the first, •viz. AM, is the same multiple of C, * that the whole of the last, viz. DL, is «ofF* B G D :et X- K C tl OF EUCLID. PROP. III. THEOR. F the first be the same multiple of the second, which the third is of the fourth ; and if of the first and third there be taken equimultiples, these shall be equimultiples, the one of the second, and the other of the fourth. Let A the first, be the same multiple of B the second, that C the third is of D the fourth ; and or A, C let the equimul- tiple- EF, GH be taken : Then EF is the same multiple of B, that GH ts of D. Because EF is the same multiple of A, that GH is of C, there arc as many magnitudes in EF equal to A, as are in GH equal to C : Let EF be di- ided into the magnitudes ■^' EK, KF, each equal to A, H* and GH into Gl,, LH, fleach equal to C : Tfie num- ber therefore of the magni- itudes EK, KF, s^all be equal to the number ot the • t- others GL, LH : And be- cause A is the same multi- ple of B, that C is of D, and that EK is equal to A, land GL to C ; therefore lEK is the same multiple E A B ofB,thatGLisofD: For I the same reason, KF is the same multiple of B, that LH is of ;D ; and so, if there be more parts m EF, GH equal to A, C : I Because, therefore, the first EK is the same multiple of the second B, which the third GL is of the fourth D, and that the fifth KF is the same multiple of the second B, which the sixth LH is of the fourth D j EF the first together with the I fifth, is the same multiple* of the second J3, which GH the » 2. 5. ' third, together with the sixth, is of the fourth D. If, there- f fore, the first, &c. Q. E. D. ft 'r C D ii8 THE ELEMENTS SceX. UookV. PROP. IV. THEOR. XF the first of four magnitudes has the same ratio to the second which the third has to the fourtli ; then any equimultiples whatever of the first and third shall have the same ratio to any equimultiples of the second and' fourth, viz. ' the equimultiple of ' the first shall have the same ratio to that of the ' second, which the equimultiple of the thii-dhas to ' that of the fourth.' Let A the first, have to B the second, the same ratio which the thrrd C has to the fourth D i and of A and C let there b« taken any equimultiples whatever E, F ; and of B and D any equi- multiples whatever G, H : Then E has the same ratio to G, which F has to H. Take of E and F any equimul- tiples whatever K, L, and of G, H, any equimultiples whatever M, N : Then, because E is the same multiple of A, that F is of C ; and of E and F have been taken equimultiples K, L ; therefore K is the same multiple of A, thatL -rr ip /» »> r' is of C*: For the same reason, M -"^ Xi A £ G is the s^me multiple of B, that N js of D : And because, as A is to L f C D IX K » Hypoth. B, so is C to D\ and of A and C have been taken certain equi- multiples K,.L : and of B and D have been taken certain equimul- tiples M, N: if therefore K be greater than M, L is greater than N : and if equal, equal ; if less, ' 5. def. 5. less<=. And K, L, are any equi- multiples whatever of E, F j and M, N any whatever of G, H : As therefore E is to G, so is*^ F to H. Therefore, if the first, &c. Q^ E, D. Cor. Likewise, if the first has the same ratio to the second, which the third has to the fourth, then also any equimultiples whatever J. j. ^ .1 £«« N. i OF EUCLID. whatever of the first and third have the same ratio to the se- Bo cond and fourth : And in like manner, the first and the third have the same ratio to any equimultiples whatever of the se- cond and fourth. Let A the first, have to B the second, the same ratio which the third C has to the fourth D, and of A and C let E and F be any equimultiples whatever ; then E is to B, as F to D. Take of E, F any equimultiples whatever K, L, and of B, Dany equimultiples whatever G, H ; then it may be demon- strated, as before, that K is the same multiple of A, that L is of C : And because A is to B, as C is to D, and of A and C certain equimultiples have been taken, viz. K and L ; and of B and D certain equimultiples G, H ; therefore, if K be greater than G, L is greater than H ; and if equal, equal ; if less, less*^ : And, K, L are any equimultiples of E, F, and G, H any what- ever of B, D; as therefore E is to B, so is F to D; And in the same way the other ca§e is demonstrated. def. 5. PROP. V. THEOR. J.F one magnitude be the same multiple of another, s« n. which a magnitude taken from the first is of a mag;- nitude taken from the other; the remainder shall be the same multiple of the remainder, that the whole is of the whole. Let the magnitude AB be the same multiple of CD, that AE taken from the first, is of CP taken from the other j the remainder EB shall be the same multiple of the remainder FD, that ;the whole AB is of the whole CD. Take AG the same multiple of FD, that AE is of CF : therefore AE is* the same mul- tiple -of CF, that EG is of CD: But AE, by the hypothesis, is the same multiple of CF, that AB is of CD : Therefore EG is the same mul- tiple of CD that AB is of CD : wherefore EG is equal to AB''. Take from them the common magnitude AE;the remainder AG is equal to -the remainder EB. Wherefore, since AE is the same multiple of CF, that AG isofFD, and that AG is equal to EB ; therefore AE is the same multiple ofCF, that EB is of FD: But AE is the same multiple of CF, 14 that A- E B M, AX.5J. 120 Book V. SceN. i Ax. 5, A K THE ELEMENTS that A B is of CD ; therefore EB is the same multiple of FD, ' that A B is of CD. Therefore, if any magnitude, &c. Q. E. D. PROP. VI. THEOR. XF two magnitudes be equimultiples of two others, and if equimultiples of these be taken from the first two, the remainders arc either equal to these others, or equimultiples of them. Let the two magnitudes AB, CD be equimultiples of the two E, F, and AG, CH taken from the first two be equimul- tiples of the same E, F ; the remainders GB, HD are either equal to E, F, or equimultiples of them. First, let GB be equal to E ; HD is ' equal to F : Make CK equal to F ; andbe- bause AG is the sam^e multiple of E, that CH is of F, and that GB is equal to E, and CK to F ; therefore AB is the same multiple of E, that KH is of F. But AB, by the hypothesis, is the same multiple of E that CD is of F ; therefore KH is the same multiple of F, that CD is of F ; wherefore KH is equal to CD* : Take away the common magnitude CH, then the remainder KC is equal to the remain- der HD : But KC is equal to F; HD therefore is equal to F. But let GB be a multiple of E ; then HD is the same multiple of F : Make CK the same multiple of F, that GB is of E: And because AG is the same mul- tiple of E, that CH is of F; and GB the same multiple of E, that CK is of F j therefore AB is the same multiple of E, that KH is of F'': But AB is the same multiple of E, that CD is of F j therefore KH is the same multiple of P', that CD is of it ; Wherefore KH is equal to CD* : Take away CH from both ; therefore the remainder KC is equal to the remainder HD : And because GB is the same multiple of E, that KC is ofF, and that KC is equal to HD ; therefore HD is the same multiple of F, that GB is of E. If therefore two magnitudes, kc. Q. E. D. C H E D E F X. A H E D E F OF EUCLID. PROP. A. THEOR. IF the first of four magnitudes has to the second, the same ratio which the third has to tiie fourth ; then, if the first be greater than the second, the third is also greater than the fourth ; and if equal, equal ; if less, less. Take any equimultiples of each of them, as the doubles of each J then, by def. 5th of this book, if the double of the first be greater than the double of the second, the double of third is greater than the double of the fourth : but, if the first be greater than the second, the double of the first is greater than the double of the second ; wherefore also the double of the third is greater than the double of the fourth ; therefore the third is greater than the fourth: In like manner, if the first be equal to the second, or less than it, the third can be proved to be equal to the fourth, or less than it. Therefore, if the first, &c. Q^ E. D. SeeN. PROP. B. THEOR. \ F four magnitudes are proportionals, they are pro- sceN. portionals also when taken inversely. If the magnitude A be to B, as C is to D, then also inverse- ly B is to A, as D to C. Take of B and D any equimultiples whatever E and F ; and of A and C any equimultiples whatever G and H. First let £ be greater than G, then G is less than E ; and because A is to B, as C is to D, and •f A and C, the first and third, G and H are equimultiples; and of B and D, the se- cond and fourth, E and F are equimulti- ples; and that G is less than E, H is also GAB *less than F ; that is, F is greater than H ; tt Q Y) if therefore E be greater than G, F is greater than H: In like manner, if E be equal to G, F may be shown to be equal (Q H ; and if less, less ; and E, F, are any equimul- tiples whatever of B and D, and G, H i . any whatever of A and C ; therefore, as B is •5.4ef.5. 122 THE ELEMENTS £oo;<v. is to A, SO IS D to C. If, then, four magnitudes, &c. ^-^^Q,E. D. PROP. C. THEOR. s«eN. J F tlie first be the same multiple of the second, or the same part of it, that the third is of the fourth; the first is to the second, as the third is to the fourth. Let the first A be the same multiple of B the second, that C the third is of the fourth D : A is to B as C is to D. Take of A and C any equimultiples what- ever E and F j and of B and D any equi- multiples whatever G and H : Then, because A is the same multiple of B that C is of D j and that E is the same multiple of A, that F is 'of C : E is the same multiple of B, that •3.5. F is of D*i therefore E and F are the same multiples of B and D : But G and H are equi- multiples of B and D: therefore, if E be a greater multiple of B than G is, F is a greater multiple of D than H is of D ; that is, if E be greater than G, F is greater than H : In like manner, if E be equal to G, or less, F is equal to H, or less than it. But E, F are equimultiples, any whatever, of A, C, and G, H, any equimultiples whatever of B, • 5. dcf. 5. D. Therefore A is to B, as C is to D^ Next, Let the first A be the same part of the second B,^ that the third C is of the fourth D : A is to B, as C is to D : For B is the same multiple of A, that D is of C: wherefore by the preceding case, B is to A, as D is to C ; and in- versely"^ A is to B as C is to D : There- fore, if the first be the same multiple, &c. Q. E. D. A B C D E G FH •B.5, A B C D OF EUCLID. PROP. D. THEOR. If the first be to the second as the third to the^^'* fourth, and if the first be a multiple or part of the second j the third is the same multiple, or the same part of the fourth. Let A be to B, as C is to D; and first let A be a multiple of B ; C is the same multiple of D. Take E equal to A, and whatever mul- tiple A or E is of B, make F the same mul- tiple of D: Then, because A is to B, as C is to D ; and of B the second, and D the fourth equimultiples have been taken E and F ; A is to E, as C to F - : But A is equal to E, therefore C is equal to F ^ : and F is the same miJtiple of D, that A is of B. A B C D Wherefore C is the same multiple of D, that A is of B. Next, Let the first A be a par£of the se- cond B ; C the third is the same part of the fourth D. Because A is to B, as C is to D ; then, inversely, B is= to A, as D to C: But A is a part of B, therefore B is a multiple of A; and, by the preceding case, D is the same multiple of C ; that is, C is the same part of D, that A is of B : Therefore, if the first, &c. Q; E. D. » Cor. 4. 5. "A. 5. See the fi- gure at th« foot of tb« preceding page- PROP. Vn. THEOR. JlLQUAL magnitudes have the same ratio to the same magnitude ; and the same has the same ratio to equal magnitudes. Let A and B be equal magnitudes, and C any-other. A and B have each of them the same ratio to C, and C has the same ratio to each of the magnitudes A and B. Take of A and B any equimultiples whatever D and E,and of 124 THE ELEMENTS SceN. E ) A B C F Book V. of C any multiple whatever F : Then, because D is the same '^'^'^'^^'^ multiple of A, that E is of B, and that A is *l. Ax. 5. equal to B ; D is* equal to E : Therefore, if ^ D be greater than F, E is greater than F ; and if equal, equal ; if less, less : And D, E are any equimultiples of A, B, and F is any mul- *5 def. 5. tiple of C. Therefore^, as A is to C, so is B to C. Likewise C has the same ratio to A, that it has to B : For, having made the same con- stru£lion, D may in like manner be shown equal to E: Therefore, if F be greater than D, it is likewise greater than E ; and ifequal, equal ; if less, less : And F is any multiple whatever of C, and D, E are any equimul- tiples whatever of A, B. Therefore C is to A, as C is to B^ Therefore, equal magni- tudes, &c. Q. E. D. PROP. Vm. THEOR. vJF unequal magnitudes, the greater has a greater ratio to the same than the less has ; and the same magnitude has .a greater ratio to the less, than it has to the greater. Let AB, B'C be unequal magnitudes, of which AB is the greater, and let D be any magnitude whatever: AB has a greater ratio to D than BC to D : And D has a greater ratio to BC than unto AB. If the magnitude which is not the greater of the tv/o AC, CB, be not less than D, take EF, FG, the doubles pf AC, CB, as in Fig. i. But if that which is not the greater of the two AC, CB be less than D (as in Fig 2 and 3.) this magnitude can be multiplied, so as to become greater than D, whether it be AC, or CB. Let it be multiplied, until it become greater than D, and let the other be multiplied as often; and let EF be the multiple thus taken of AC, and FG the same multiple of CB : Therefore EF and FG are each of them greater than D: And E Fig. A I. G L K. H D OF EUCLID. D : And in every one of the cases, take H the double of D, K its triple, and so on, till the multiple of D be that which first becomes greater than FG : Let L be that multiple of D which is first greater than FG, and K the multiple of D which is next less than L. Then, because L is the multiple of P, which is the first that becomes greater than FG, the next preceding multiple K is not greater than EG j that is, FG is not less than K : And since EF is the same multiple of AC, that FG is of CB j FG is the same multiple of CB ; that EG is of AB* ; wherefore EG and FG are equimultiples of AB and CB : And it was shown, that FG was not less than K, and, by the construe- Fig. 2. Fig. 3. tion,EF, is greater than Y.\ E D ; therefore the whole EG is greaterthanKand F D together : But K, to- gether with D, is equal to L ; therefore EG is greater than L; but FG is not greater than L ; and EG, FG are equi- multiples of AB, BC, Q Y^ C and L is a multiple of D; therefore b AB has L K H D to D a greater ratio than BC has to D. L K D Also D has to BC a greater ratio than it lus to AB : For, having made the same con- stru£bion, it may be shown, in like manner, that L is greater than FG, but that it is not greater than EG : and L is a multiple of D ; and FG, EG are equimultiples of CB, AB ; therefore D has to CB a greater ratio'' than it has to AB. Wherefore, •f unequal magnitudes, &c. Q^ E. D. A C 1.5. det SwN. THE ELEMENTS PROP. IX. THEOR. IVaagnitudes which have the same ratio to the same iliagnitude are equal to one another; and those to which the same magnitude has the same ratiQ are equal to one another. Let A, B have each of them the same ratio to C : A is equal to B : for, if they are not equal, one of them is greater than the other ; let A be the greater ; then, by what was shown in the preceding proposition, there are some equimultiples of A and B, and some multiple of C such, that the multiple of A is greater than the multiple of C, but the multiple of B is not greater than that of C. Let such multiples be taken, and let I), E, be the equimultiples of A, B, and F the multiple of C, so that D may be greater than F, and E not greater than F : But, beeause A is to C, as B is to C, and of A, B, are taken equimultiples D, E, and of C is taken a multiple F j and that Vv D is greater than F ; E shall also be greater - 5. def. 5. than F^ ; but E is not greater than P" ; ^ which is impossible J A therefore and B are not unequal ; that is, they are equal. Next, let C have the same ratio to each of the magnitudes A and B ; A is equal to B : For, if they are not, one of them is ■.> .greater than the other ; let A be the great- er ; therefore, as was shown in Prop. 8th, there is some multiple F of C, and some equimultiples E and D, of B and A such, that F is greater than E, and not greater than D ; but because C is to B, as C is to A, and that F, the multiple of the first, is greater than E, the multiple of the second ; F the multiple of the third, is greater than D, the multiple of the fourth' : But F is not greater than D, which is impossible. Tliere- fore A is equal to B. Wherefore, magnitudes which, &c. Q,E.D. OF EUCLID. PROP. X. THEOR. X HAT magnitude which lias a greater ratio than s«ex another has unto the same magnitude, is the s:reat- er of the two : And that magnitude to which the same has a greater ratio than it has unto another magnitude is the lesser of the two. Let A have to C a greater ratio than B has to C ; A is great- er than B : For, because A has a greater ratio to C, than B has to C, there are* some equimultiples of A and B, and some *7. def. 5. multiple of C such, that the multiple of A is greater than the multiple of C, but the multiple of B is not greater than it : Let them be taken, and let D, E be equir ul- tiples of A, B, and F a multiple of C such, that D is greater than F, but E is not J) greater than F : Therefore D is greater than E : And, because D and E are equi- multiples of A and B, and D is greater A than E ; therefore A is^ greater than B. Next, let C have a greater ratio to B ^1 V **.Ax.5. than it has to A ; B is less than A : For* there is some multiple F of C, and some equimultiples E and D of B and A such, B that F is greater than E, but is not greater than D : E therefore is less than D ; and because E and D are equimultiples of B and A, therefore B is'' less than A. That magnitude, there- fore, &c. Q. E. D. PROP. XL THEOR. Jtv ATIOS that are the same to the same ratio, are the same to one another. Let A be to B as C is to D; and as C to D, so let E be to F J A is to B, as E to F. Take of A, C, E, any equimultiples whatever G, H, Kj and of B, D, F, any equimultiples whatever L,M, N. Therefore, since A is to B, as C to D, and G, H are taken equmultiples of A,C, 128 ■ THE ELEMENTS OOK V » :). def. 5. $ , C, and L, M, of B,D ; if G be greater than L, H is greater than Mj and if equal, equal; and if less, less\ Again, be- cause C is to D, as E is to F, andH,Kare taken equimultiples of C, E ; and M, N, of D, F ; if H be greater than M, K is greater than N ; and if equal, equal ; and if less, less : But if G H K A C E ■ B I> F L M N G be greater than L, it has been shown that H is greater than M, and if equal, equal ; and if less, less ; therefore, if G be greater than L, K is greater than N ; and if equal, equal ; and if less, less : And G, K are any equimultiples whatever of A, E ; and L, N any whatever of B, F : Therefore, as A is to B, so is E to F\ Wherefore, ratios that, Sec. Q. E. D. PROP. XII. THEOR. JlF any number of magnitudes be proportionals, as one of the antecedents is to its consequent, so shall all the antecedents taken together be to all the con- sequents. Let any number of magnitudes A, B, C, D, E, F, be pro- portionals ; that is, as A is to B, so C to D, and E to F : As A is to B, so shall A, C, E together be to B, D, F together. Take of A, C, E any equimultiples whatever G, H, K ; G H K A C F B D— , F L M-^^ N- and of B, D, F any equimultiples whatever L, M, N : Then, because A is to B, as C is to Djand as E to F ; and that G, H, K OF EUCLID. 129 K are equimultiples of A, C, E, and L, M, N, equimultiples of BookV. B, D, F ; if G be greater than L, H is greater than M, and K iS.Def. 5. greater than N ; and if equal, equal ; and if less, less*. Where- fore, if G be greater than L, then G, H, K together are greater than L, M, N together j and \i equal, equal ; and if less, less. And G, and G, H, K, together are any equimultiples of A, and A, C, E togetner ; because if there oe any number of magni- tudes equimultiples of as many, each of each, whatever mul- tiple of one of them is of its part, the same multiple is the % whole of the whole": For the same reason L, and L, M, N ^" ^' are any equimultiples of B, and B, D, F : As therefore A is to B, so are A, C, E, tosether to B, D, F together. Where^ fore, if any number, &c. Q. E. D. PROP. XIII. THEOR. JLF the first has to the second the same ratio which ^^'' the third has to the fourth, but the third to the fourth a greater ratio than the fifth has to the sixth; the first shall also have to the second a greater ra- tio than the fifth has to the sixth. Let A the first, have the same ratio to B the second, which C the third, has to D the fourth, but C the third, to D the fourth, a greater ratio than E the fifth, to F the sixth : Also the first A shall have to the second B a greater ratio than the fifth E to the sixth F. Because C has a greater ratio t« D, than E to F, there are some equimultiples of C and E, and some of D and F such, that the multiple of C is greater than the multiple of D, but M G H B D F N K L the multiple of E is not greater than the multiple of F» : Let • T. drf. 5. Isuch be taken, and of C, E let G, H be equimultiples, and K, |iL equimultiples of D, F, so that G be greater than K, but H aot greater than L ; and whatever multiple G is of G, take M :he same multiple of A ; and whatever multiple K isof D, take M the same multiple of B : Then, because A is to B, as C to K D, and I 1 130 THE ELEMENTS '5.Def,5. Def. 5. Book V. J)^ jfjd of A and C, M and G are equimultiples ; And of B and D, N and K are equimultiples ; if M be greater than N, G is greater than K ; and if equal, equal ; and if less, less'' ; but G is greater than K, therefore M is greater than N : But H is not greater than L ; and M, H are equimultiples of A, E ; and N, L equimultiples of B, F : Therefore A has a greater ratio to B, than E has to F^ Wherefore, if the first, &c. Q^ E. D. Cor. And if the first have a greater ratio to the second, than the third has to the fourth, but the third the same ratio to the fourth, which the fifth has to the sixth j it may be demon- strated, in like manner, that the first has a greater ratio to the second, than the fifth has to the sixth. PROP. XIV. THEOR. See N. J[f the first has to the second, the same ratio which the third has to the fourth ; then, if the first be greater than the third, the second shall be greater than the fourth ; and if equal, equal ; and if less, less. Let the first A have to the second B, the same ratio which the third C, has to the fourth D j if A be greater than C, B is greater than D. Because A is greater than C, and B is any other magnitude^ « 1. 5. A has to B a greater ratio than C to B'' ; But, as A is to B, sq 3 A B C D A B C D A B C D is C to D i therefore also C has to D a greater ratio than C has to B^ But of two magnitudes, that to v^ich the same has the greater ratio is the lesser^ Wherefore D is less than B ; that is, B is greater than D. Secondly, if A be equal to C, B is equal to D : For A is to B, as C, that is A, to D : B therefore is equal to D*". Thirdly, if A be less than C, B shall be less than D : For C is greater than A, and because C is to D, as A is to B, D is greater than B, by the first case ; wherefore B is less than Di Therefore, if the first, &c. Q.. E. D. » 13. 5. « 10. 5. *9.5. OF EUCLID. PROP. XV. THEOR. IVIagnitudes have the same ratio to one ano- ther which their equimultiples have- Let AB be the same multiple of C, that DE is of F j C is f F, as AB to DE. Because AB is the same multiple of C, thatDE is of F; there areas many magnitudes in AB equal to C, as there are in DE equal to F : Let AB be divided into magnitudes, each equal to C, tx viz. AG, GH, HB ; and DE into magni- tudes, each equal to F, viz. DK, KL, LE : Then the number of the first AG, GH, H B, K shall be equal to the number of the last DK, KL, LE : And because AG, GH, HB are all equal, and that DK, KL, LE, are also equal to one another ; therefore AG iS to DK, as GH to KL, and as HB to LE M B C E F And as one of the antecedents to its conse- quent, so are all the antecedents together to all the consequents together'' ; wherefore, as AG is to DK, so is AB to DE : But " ^2- 5. AG is equal to C, and DK to F : Therefore, as C is to F, so is AB to DE. Therefore magnitudes, &c. Q^ E. D. H- PROP. XVL THEOR. xFfour magnitudes of the same kind be proportion- als, they shall also be proportionals when taken al- ternately. Let the four magnitudes A, B, C, D, be proportionals, viz. as A to B, so C to D : They shsill also be proportionals when taken alternately ; that is, A is to C, as B to D. , Take of A and B any equimultiples whatever E and F ; and •f C and D take any equimultiples whatever G and H : and K 2 because THE ELEMENTS _^^^ because E is the same multiple of A, that F is of B, and that M3.5. magnitudes have the same ratio to one another which their equimultiples have^j therefore A is to B, as E is to F : But as A is to B, so is C to »il. 5. D : Wherefore as C E" — G is to D, so *> is E to p F: Again, because ^ ^ G, H are equimul- i> T) tiples of C, D, as C is to D, so is G to F H H» ; but as C is to D, so is E to F. Wherefore, as E is to F, so is G to H^ But when four magnitudes are proportionals, if the first be greater than the third, the second shall be greater than the fourth ; and ' U. 5. If equal, equal j if less, less"^. Whtrefore, if E be greater than G, F likewise is greater than H : and if equal,, equal ; if less, less : And E, F arc any equimultiples whatever of A, B ; and ^ G, H any whatever of C, D. Therefore A is to C as B to *5. Def. 5. Dd. If then four magnitudes, &c. Q, E. D. SccN. PROP. XVII. THEOR. JLF magnitudes, taken jointly,' be proportionals, they shall also be proportionals when taken sepa- rately; that is, if two magnitudes together have to one of them the same ratio which two others have to one of these, the remaining one of the first two shall have to the other the same ratio which the remaining one of the last two has to the other of these. Let AB, BE, CD, DF be the magnitudes taken jointly which are proportionals'; that is, as AB to BE, so is CD to DF ; they shall also be proportionals taken separately, viz. as AE to EB, so CF to FD. Take of AE, EB, CF, FD any equimultiples whatever GH, HK, LM,MN; and again, of EB, FD take any equimultiples whatever KX, NP : And because GH is the same multiple of AE, that HK is of E B, wherefore GH is the same multiple* of AE, that GK is of AB : But GH is the same multiple of AE, that LMi^ ofCFj wherefore GK is the same multiple of AB, I that OF EUCLID. ^33 X K *2. 5. that LVI is of CF. Again, because LM is the same mukrple of Book v CF, that MN is of FD; therefore LM is the same multiple^'^'^^ of CF, than LN is of CD : But LM was shown to be the same multiple of CF, that GK is of AB ; GK therefore is the same multiple of A H, that LN is of CD ; that is, GK, LN areequi- multiples of AB, CD. Next, because HK is the same multiple ofEB, thatMNisofFD; and that KX is also the same multiple of EB, that NP is of FD; therefore HX is the same multiple *'of EB, that MP is of FD. And because AB is to BE, as CD is to DF, and that of AB and CD, GK. and LN are equimul- tiples, and of EB and FD, HX and MP are equimultiples ; if GK be greater than HX, then LN is greater than MP ; and if equal, equal ; and if less, Iess<^ ; But if H GH be greater than KX, by adding the common part HK to both, GK is greater than HX ; wherefore also LN is greater than MP ; and by talcing away MN from both, LM is greater than NP : There- fore, if GH be greater than KX, LM is greater than NP. In like manner it may be demonstrated, that if GH be equal to KX, LM likewise is equal to NP; and if less, less : And GH, LM are any equi- multiples whatever of AE, CF, and KX, NP are any what- ever of EB, FD. Therefore^ as AE is to EB, so is CF to ¥D. If then magnitudes, &c. Q^ E. D. B E D F N M' 5. De£ 5. G A C L PROP. XVIII. THEOR. J-F magnitudes, taken separately, be proportionals, they shall also be proportionals "when taken jointly, that is, if the first be to the second, as the third to the fourth, the first and second together shall be to the second, as the third and fourth together to the fourth. LetAE, EB, CF, FD be proportionals ; that is, as AE to EB, so is CF to FD ; they shall also be proportionals whcQ- taken jointly ; that Is, as AB to BE, so CD to DF. Take of AB, BE, CD, DF any equimultiples whatever GH, HK, LM, MN : and again, of BE, DF, take any what- ever equimultiples KG, NP: And because KO, NP are K 3 equimultiples. Seeh. 13+ THE ELEMENTS K «S.5. M P- N ^°°^- equimultiples of BE, DF ; and thatKH, NMare equimultiples ^"^""^^^ likewise of BE, DF, if KO, the multiplcof BE, be greater than KH, which is a multipleof the same BE, NP, likewise the mul- tiple of DF, shall be greater than MN, the multiple of the same DF ; and if IT KO be equal to KH, MP shall be cq«al to NM ; and if less, less. First, Let KO not be greater than KH, therefore NP is not greater than NM : And because GH, HK, are equi- multiples of AB, BE, and that AB is greater than BE, therefore GH is « 5. Ax. 5. greater* than HK j but KO is not greater than KH, wherefore GH is greater than KO. In like manner it B may be shown, that LM is greater than NP.. Therefore, if KO be not greater E than KH, then GH, the multiple of AB, is always greater than KO, the ^ multiple of BE ; and likewise LM, the multiple of CD, greater than NP, the multiple of DF. Next, Let KO be greater than KH : therefore, as has been shown, NP is greater than NM : And because the whole jH is the same multiple of the whole AB, that HK is of BE, the remainder GK is the same multiple of *5.5. the remainder AE that GH is of AB^: q which is the samethat LM is of CD. In like manner, because LM is the K same multiple of CD, that MN is of DF, the remainder LN is the same multiple of the remainder CF, that the whole LM is of the whole CD'' : K But it was shown that LM is the same -n " multiple of CD, that GK is of AE j D" therefore GK is the same multiple of E AE, thatLN isof CF; that is, GI^, LN are equimultiples of AE, CF : And because KO, NP are equimul- G A tiples of BE, DF, if from KO, NP, there be taken KH, NM, which are likewise equimultiples of BE, DF, the remainders HO, MP are either equal to B?2, DF, or equimultiples of them ^ First, let HO, MP, be equal to BE, DF j and because AE is to EB, as CF to FD, and that P M N OF EUCLID. that GK, LN are equimultiples of AE, CF ; GK shall be to EB, as LN to FD' : But HO is equal to EB, and MP to FD ; wherefore GK is to HO, as LN to MP. If therefore GK be greater than HO, LN is greater than M P ; and if equal, equal ; and if less^ less. But let HO, MP be equimultiples of EB, FD ; and because AE is to EB, as CF to FD, and that of AE, CF are taken equimultiples GK LN ; andofEB,FD,theequimultiplesHO, MP ; if GK be greater than HO, LN is greater than MP ; and if equal equal ; and if less, less*^ ; which was likewise shown in the preceding case. If thereforeGH be greater than KO, taking KH from both, GK is greater than HO ; wherefore also LN is greater than MP ; and consequently, adding NM to both, LM is greater than NP : Therefore, if GH be greater than KO, LM is greater than NP. In like manner it maybe shown, that if GH be equal to KO, LM is equal to NP ; and if less, less. And in the case in which KO is not greater than KH, it has been shown that GH is always greater than KO, and likewise LM than NP : But GH, LM are any equimultiples of AB, CD, and KO, NP are any whatever of BE, DF ; therefore*^, a«; AB is to BE, so is CD to DF. If then magnitudes, &c. Q^ E. D. Book V. ^ Cor, 45. = A. 5. H G B D F A C M N f=; 5. D«f. 5. PROP. XIX. THEOR. JLF a whole magnitude be to a whole, as a magui- See n. tude taken from the first, is to a magnitude taken from the other; the remainder shall be to the re- mainder, as the whole to the whole. Let the whole AB, be to the whole CD, as AE, a magni- tude taken from AB, to CF, a magnitude taken from CDj the remainder EB shall be to the remainder FD, as the whole AB to the whole CD. Because AB is to CD, as AE to CF : likewise, alternately*, » le 5, K4 BA - '36 Book V. " 17. 5. THE ELEMENTS BA is to AE, as DC Is te CF : and because, if magnitudes, taken jointly, be proportionals, they are also proportionals^ when taken separately j therefore, as BE is to EA, so is DF to FC, and alternately, as BE is to DF, so is EA to FC : But, as AE toCF, so by the hypothesis, is AB to CD J therefore also BE, the remainder, shall be to the remainder DF, as the whole AB to the whole CD : Wherefore, if the whole, &c. Q. E. D. Cor. If the \V^hole be to the whole, as a mag- nitude taken from the first,is to a magnitude taken B D from the other ; the remainder likewise is to the remainder ; as the magnitude taken from the first to that taken from the other ; The demonstration is contained in the pre- ceding. PROP. E. THEOR. 1 F four magnitudes be proportionals, they are also proportionals by conversion ; that is, the first is to its excess above the second, as the third to its ex* cess above the fourth. » 17. 5. •>B. 5. « IS. 5. Let AB be to.BE, as CD to DF j then BA is to AE, as DC to CF. Because AB is to BE, as CD to DF, by divi- sion*, AE is to EB, as CF to FD ; and by in- version'', BE is to EA, as DF to FC. Where- fore, by composition", BA is to AE, as DC is to CF : If, therefore, lour, &c. Q^ E. D. E B n SceN. PROP. XX. THEOR. If there be three magnitudes, and other three, which, taken two and two, have the same ratio; if" the first be greater than the third, the fourth shall be greater than the sixth ; and if equal, equal ; and if less, less. Let OF EUCLID. ^37 D E F Let A, B, C be three magnitudes, and D, E, F other three, BookV^ which, taken two and two, have the same ratio, viz. as A is to ^■^""'''^^ B, so is D to E ; and as B to C, so is E to F. If A be greater than C, D shall be greater than F : and, if equal, equal ; and if less, less. Because A is greater than C, and B is any other magnitude, and that the greater has to the same magnitude a greater ratio than the less has to it^ ; therefore A has to B a greater ratio than C has to B : Bui as D is to E, so is A to A B B ; therefore'' D has to E a greater ratio than C to B ; and because B is to C, as E to F, by inversion, C is to B, as F is to E : and D was shewn to have to E a greater ratio than C to B ; therefore D has to E a greater ratio than F to E^ But the magnitude which has a greater ratio than another to the same magnitude, is the greater of the two** : D is therefore greater than F. Secondly, Let A be equal to C ; D shall be equal to F cause A and C are equal to one another, A is to B, as C is to B^ : But A is to B, as D to E ; and C is to B, as F to E J wherefore D is to E, as F to Ef ; and therefore D is equal to Fs. Next, Let A be less than C ; D shallbe less than F: For C is greater ^ than A, and, as was shown in the first case, C is to B, as F to E, and in like manner B is to A, as E to D ; therefore F is greater than D, by the first case ; aad therefore D is less than F. Therefore, if there be three, &c. Q. E. D. Be- A B C Jb A D B C E F 8.5. * 13.5. "=Cor. 13.5. 10. 5. 7,5. ni.5. '9.5. PROP. XXL THEOR. IF there be three magnitudes, and other three, SceN. whicli have the same ratio taken two and two, but in a cross order; if the first magnitude be greater than the third, the fourth shall be greater than the sixth ; and if equal, equal ; and if less, less. Let ^3? THE ELEMENTS •8. 5. » 13. 5 'Cor. 13. 5 « 10.5 « 7. 5. fll. 5 e 9.5. BookV. Let A, B, C be three magnitudes, and D, E, F other three, ^'^''^'^^ which have the same ratio, taken two and two, but in a cross order, viz. as A is to B, so is E to F, and as B is to C, so is D to E. If A be greater than C, D shall be greater than F ; and if equal, equal j and if less, less. Because A is greater than C, and B is ariy other magnitude, A has to B a greater ratio* than C has to B : But as E to F, so is A to B : therefore^ E has to F a greater ratio than C to x B : And because B is to C. as D to E, by inver- ^ IJ C sion, C is to B, as E to D : And E was' shown to have to F a greater ratio than C to B ; there- - J fore E has to F a greater ratio than E to 0*= ; but the magnitude to wliich the same has a greater ratio than it has to another, is the lesser of the two** : F therefore is less than D j that is, D is greater than F. Secondly, Let A be equal to C ; D shall be equal to F. Be- cause A and C are equal, A is^ to B, as C is to B : But A is to B, as E to F ; and G is to B, as E to D ; wherefore E is to F as E to D^ ; and therefore D is equal to F^. ■ Next, Let A be" less than C i D shall be less than F : For C is greater than A, and, as was shown, C is to B, as E to D, and in like manner B is to A, D as F to E ; therefore F is greater than D, by case first; and there- fore, D is less than F. There- fore, if there be three, &c. Q, E. D. - A B C. A JB C D E F ' PROP. XXn. THEOR. SteN. If there be any number of magnitudes, and as many others, which, taken two and two in order, have the same ratio ; the first shall liave to the last of the first magnitudes the same ratio which tlie first of the others has to the last. N. B. This is usually cited hi/ the xvords " e.v ccquali,'" oi', "e.i" cequo,'' First, OF EUCLID. ^39 D E F H L N »4.5. First, Let there be three magnitudes A, B, C, and as many ^°^^" others D, E, F, which, taken two and two, have the same ^■^'v^' ratio i that is, such that A is to B as D to E j and as B is to C, so is E to F ; A shall be to C, as D to F. Take of A and D any equimultiples whatever G and H ; and of B and E any equimultiples whatever K and L ; and of C and' F any whatever M and N : Then, because A is to B, as D to E, and that G, H are equimultiples of A, D, and K, L equimultiples of B, A "R r* E ; as G is to K, so is* H to L : "^ For the same reason, K is to M, n rr vr as L to N ; and because there are T" three magnitudes G, K, M, and other three H, L, N, which, two and two, have the same ratio ; if G be greater than M, H is greater than N ; and if equal, equal ; and if less, less"; and G, H are any equimultiples whatever of A, D, and M, N are any equimul- tiples whatever of C, F : Therefore*^, as A is to F. Next, I^t there be four magnitudes. A, B, C, D, and other four E, F, G, H, which, twoand two, have the same ratio, viz. as A is to B, so is E to F; and A. B. C. D. i as B to C, so F to G : and as C to D, so G to E. F. G. H. i H : A shall be to D, as E to H. ' Because A, B, C, are three magnitudes, and E, F, G, other three, which, taken two and two, have the same ratio ; by the foregoing case, A is to C, as E to G : But C is to D, as G is to H ; wherefore again, by the first case, A is to D, as E to H ; a-.id so on, whatever be the number of magnitudes. Therefore, if there be any number. Sec. Q^ E. D. "20.5. to C, so is D ' 5. Dcf. S. THE ELEMENTS PROP. XXIII. THEOR. SecN. 15.5. bll.5. 4.5. *21.5. J. F there be any number of magnitudes, and as many others, which, taken two and two, in a cross order, have the same ratio ; the first shall have to the last of the first magnitudes the same ratio which the first of the others has to the last. N. B. T/iis u- usually cited hy the words, "c.r cequali in pi^o- portione perturhata ;" or, *' ex (Zquo perturbato.''^ First, Let there be three magnitudes A, B, C, and other three D, K, F, which, taken two and two, in a cross order, have the same ratio, that is, such that A is to B, as E to F ; and as B is to C, so is D to E: A is to C, as D to F. Take of A, B, D any equimultiples whatever G, H, K ; and of C, E, F any equimultiples whatever L, M, N : And because G, H are equimultiples of A, B, and that magnitudes have the same ratio which their equimul- tiple have* ; as A' is to B, so is G to H : And for the same rea- son, as E is to F, so is M to N : But as A is to B, so is E to F i as therefore G is to H, so is M to N^ And because as B is to C, ^ ^ ^ K. M ? so is D to E, and that H, K are equimultiples of B, D, and L, M of C, E ; as H is to L, so is"^ K to M : And it has been shown that G is to H, as M to N : Then, because there are three magni- tudes G, H, L, and other three K, M, N, which have the same ratio taken two and two in a cross order ; if G be greater than L, K is greater than N ; and if equal, equal ; and if less, Icss^ ; and, G, K, are any equimultiples whatever of A, D; and L> N any whatever of C, F: as, therefore, A is to C, so is D to F, Next, ABC B f OF EUCLID. Next, Let there be four magnitudes, A, B, C, D, and other four E, F, G, H, which, taken two and two, in a cross order, have the same ratio, viz. A to B, as G to H ; B to C, as F to G ; and G to D, as E to F : A is to D, as E to H. Because A, B, C, are three magnitudes, and F, G, H other three, which taken two and two, in a cross order, have the same ratio ; by the first case, A is to C, as F to H : But C is to D, as E is to F; wherefore again, by the first case, A is to D, as E to H : And so on, whatever be the number of mag- nitudes. Therefore, if there be any number, &c. Q. E. D. PROP. XXIV. THEOR. 1 F the nrst has to the second the same ratio which the third has to the fourth ; and the fifth to the se- cond, the same ratio which the sixth has to the fourth ; the first and fifth together shall have to the second, the same ratio which the third and sixth togetlier have to the fourth. Let AB the first, have to C the second, the same ratio which DE the third, has to F the fourth j and let iSG the fifth have to C the second, the same ratio which EH the sixth, has to F the fourth ; AG, the G first and fifth together, shall have to C the second, the same ratio' which DH, the third and sixth together, has to F the fourth. Because BG is to C, as EH to F ; by in- version, C is to BG, as F to EH : And be- cause, as AB is to C, so is DE to F ; and as C to BG, so F to EH ; ex a?quali% AB is to BG, as DE to EH : And be- cause these magnitudes are proportionals, they shall likewise be proportionals when taken jointly*" j as therefore AG is to GB, so is DH to HE ; but as GB to C, so is HE to F. B H A C D F SeeN*. '22,5. »I3. Th( re- fore ex aequalis as AG is to C, so is DH to F. .Wherefore ifthefirst, &c. Q. E. D. Cor. I. If the same hypothesis be made as in the proposi- tion, the excess of the first and fit'th shall be to the second, as the THE ELEMENTS the excess of the third and sixth to the fourth : The demon- stration of this is the same with that of the proposition, if di- vision be used instead of composition. Cor. 2 The proposition holds true of two ranks of magni-' tudes, whatever be their number, of whic"h each of the nrst rank has to the second magnitude the same ratio that the cor- responding one of the second rank has to a fourth magnitude ', as is manifest. PROP. XXV. THEOR. JLF four magnitudes of the same kind are prportion- als, the greatest and least of them together arc greater than the other two together. Let the four magnitudes AB, CD, E, F be proportionals, viz. AB to CD, as E to F i and let AB be the greatest of them, • A. & I'i. ^'^^ consequently F the leasts AB, together with F, are 5. greater than CD, together with E. Take AG equal to E, and CH equal to F : Then because as AB is to CD, so is E to F, and that AG is equal to E, and CH equal to F, AB Is to CD, as AG to CH. And because AB the whole, is to the whole CD, as AG is to CH, likewise the p remainder GB shall be to the remainder "A. 5 19.5. HD, as the whole AB is to the whole'' CD : But AB is greater than CD, there- fore"^ "GB is greater than HD : And be- cause AG is equal to E, and CH to F j AG and F together are equal to CH and E together. If therefore to the unequal x n rr t magnitudes GB, HD, of which GB is -^ ^ ^ ^ the greater, there be added equal magnitudes, viz. to GB the two AG and F, and CH and E to HD j AJB'and F together are greater than CD and E. Therefore, if four magnitudes, &c. Q. E. D. PROP. F. THEOR. S<«N. XxATIOS which are compounded of the same ra- tios, are the same with one another. Let OF EUCLID. 143 Let A be to B, as D to E ; and B to C, as E to F : The ni- ^-^ok v. tio which is compounded of the ratios of A ^-"v^^ to B, ana B to C, which by the definition of compound ratio, is the ratio of A to C,^ is the same with the ratio of D to F, which by the same definition, is compounded of the ratios of D to E, and E to F. Because there are three magnitudes A, B, C, and three others D, E, F, which, taken two and two, in order, have the same ratio ; ex aequali A is to C, as D to F*. Next, Let A be to B, as E to F, and B to C, as D to E ; there- for?, ex aquali in proportion perturbatct^^ A is to C, as D to F ; that is, the ratio of A to C, which is compounded of the ratios of A to B, and B to C, is the same with the ratio of D to F, which is compounded of the ra- tios of D to E, and E to F : And in like manner the proposi- tion may be demonstrated, whatever be the number of ratios in either case. '22 5. b23. 5. A. B. C. D. E. F. PROP. G. THEQR. IF several ratios be the same with several ratios, c v each to each ; the ratio which is compounded of ratios which are the same with the first ratios, each to each, is the same with the ratio compounded of ratios w hich are the same with the other ratios, each to each. Let A be to B, as E to F ; and C to D, as G to H : A.nd let A be to B, as K CO L ; and C to D, as L to M : Then the ra, tio of K to M, by the definition - — . of compound ratio, is compound- 1 A. B C D K L M E. F. G. H. N. O. P. ed of the ratios of K to L, and L to M, which are the same with the ratios of A to B, and C to D : And as E to F, so let N be to O ; and as G to H, so let O be to P J then the j-aiio of N" to P, is compounded of the ratios of N to O, and O to P, which are the same with the ratios of E to F, and G to H : And it is to be shown that the ratio of K to M, is the same with the ratio of N to P, or that K is to M, asNtoP. Because K is to L, as (A to B, that is, as E to F, that is, as N toOi and as L to M) so is (C to D, and so is G to H, and SeeN. THE ELEMENTS and so is'O to P:) Ex xquali=' K is to M, as N to P. Thetc- »22. 5. fore, if several ratios, &c. Q. E. D. PROP. H. THEOR. IF a ratio compounded of several ratios be the same with a ratio compounded of any other ratios, and if one of the first ratios, or a ratio compoun- ded of any of the first, be the same with one of the last ratios, or with the ratio compounded of any of the last ; then the ratio compounded of the re- maining ratios of the firsts or the remaining ratio of the first, if but one remain, is tlie same with the ratio compounded of those remaining of the last, or with the remaining ratio of the last. Let the first ratios be those of A to B, B to G> C to D, D to E,^and E to F ; and let the other ratios be those of G to H, H to K, K to L, and L to M ; Also, let the ratio of A to o?coml"°" F, which is compounded of^ the first A. B. C. D. E. F. G. H. K. L. M. pounded ratios,be the same with theratioof G "''°* to M, which is corapounded of the other ratios: And besides, let the ra- tio of A to D, which is compounded of the ratios of A to B, B to C, C to D, be the same with the ratio of G to K, which is compounded of the ratios of G to H, and H to K : Then the ratio compounded of the remaining first ratios, to wit, of the ratips of D to E, and E to F, which compounded ratio is the ratio of D to F, is the same with the ratio of K to M, which is compounded of the remaining ratios of K to L, and L to M of the other ratios. Because^ by the hypothesis, A is to D, as G to K, by in- b B 5. version'', D is to A, as K to G ; and as A is to F, so is G to ' 22. 5. M J therefore^ ex aequali, D is to F, as K to M. If there- fore a ratio which is, &c, Q^ E. D. O^ EUCLia I4S BcoxV. PROP. K. THEOR. J.F there be any number of ratios, and any number see n. of other ratios such, that the ratio compounded of ratios which are the same with the first ratios, each to each, is the same with the ratio compounded of ratios "which are the same, each to each, with the last ratios; and if cne of the first ratios, or the ratio which is ' compounded of ratios which are the same with several of the fii-st ratios, each to each, be the same with one of the last ratios, or with the ratio compounded of ra- tios which are the same, each to each, with several of the last ratios : Then the ratio compounded of ratios which arc the same with -the remaining ratios of the first, each to each, or the remaining ratio of the first, if but one remain ; is the same with the ratio com- pounded of ratios which are the same with those re- maining of the last, each to each, or with the re- maining ratio of the last. Let the ratios of A to B, C to D, E to F, be the first ratios ; and the ratios of G to H, K to L, .\1 to N, O to P, Q to R, be the other ratios : And let A be to B, as S to T ; and C to D, as T to V, and E to F, as V to X : Therefore, by the deii- nition of compound ratio, the ratio of S to X is compounded h, k, 1. A, Bi C, D; E, F. S, T, V. X. G,H;K, LjM,NjO,PiQjR. Y, Z,a, b, c,d. e, f, g. . m, n, o, p. t)f the ratios of S to T, T to V, and V to X, which are the Isame with the ratios of A to B, C to D, E to F, each' to each : I Also, as G to H, so let Y be to Z ; and K to L, as Z to a ; M Ito N, as a to b, O to P, as b to c j and Q^to R, as c to d : ITherefore, by the same definition, the ratio of Y to d is com- Ipounded of the ratios of Y to Z, Z to a, a to b, b to c, and I L c to 146 THE ELEMENTS OF EUCLID. BookV^ q to d, which are the same, each to each, with the ratios of G ''"^^'^^ to H, K to L, M to N, O to P, ancl Qjo R : Therefore, by the hypothesis, S is to X, as Y to d ; Also, let the ratio of A to B, , that is, the ratio of S ,to T, which is one of the first ratios, be the same with the ratio of e to g, which is compounded of the ratios of e to f, and f to s;, which, by the hypothesis, are the same with the ratios of G to H, and K to^ L,two of the other ratios j and let the ratio of h to 1 be that v/hich is compounded of the ratios of h to k, and k to 1, which are the same with the remaining first ratios, viz. of C to D, and E to F 5 also, let the ratio of m to p, be that which is compounded of the ratios of m to n, n to o, and o to p, which are the same, each to each, with the remaining other ratios, viz. of iM to N, O to P, and Q to R : Then the ratio of h to 1 is the same with the ratio of m to p, or h is to 1, as m to p. U.5. h, k, i. A, B ; C, D i E, F. S,T,V,X. G, H; K, L; M, N J 0, P; Q, P« Y, Z, a, b. c, d. e, f, g. m, n, 0, p. - Because c is,' to f, as (G to H, that I?, as) Y to Z j and f is to g, as (K to L, that is, as) Z to a ; therefore, ex icquali, e is to g, as Y to a : And by the hypothesis, A is to B, that is, S to X, as e to g ; wherefore S is to T, as Y to a ; and, by inversion, T is to S, as a to Y ; and S is to X, as Y to d ; therefore, ex sequali, T is to X, as a to d : Also, because h is to k as (C to D, that is, as) T to V j and k is to 1, as (E to F, that is, as) V to X ; therefore, excequali, h is to 1, ?.s T to X : In like manner, it may be demonstrated, that m is to p, as a to d : And it has been shown, that T is to X, as a to d ; therefore" h is to 1, as m to p. Q. E. D. The propositions G and K are usually, for the sake of bre- vity, expressed in the same terms with propositions F and H: And therefore it was proper to show the true meaning of them when they are so expressed ; especially smce they are very frequently made use of by geometers. [ 147 ] THE ELEMENTS EUCLID. BOOK VI. DEFINITIONS. I. Similar re£tilineal figures are those which have their several angles equal, each to each, and the sides about the equal angles proportionals. Book VI. ^ II. "Reciprocal figures, viz. triangles and parallelograms, are See N. " such as have their sides about two of their angles propor- *' tionals in such manner: that a side of the first figure is to *' a side of the other, as the remaining side of this other is ** to the remaining side of the first." III. A straight line is said to be cut in extreme and mean ratio, when the whole is to the greater segment, as the greater segment is to the less.^ IV. The altitude of any figure is the straight line drawn from its vertex perpendicular to the base. THE ELEMENTS PROP. I. THEOR. seeN. j^ RiANGLES and parallelograms of thc saiiic al- titude are one to another as their bases. Let the triangles ABC, ACD, and the parallelograms EC, CF have the same ali'tude, viz. the perpendicular drawn from the point A to BD : Then, as the base BC, is to the base CD, so is the triangle ABC to the triangle ACD, and the paral- 'lelogram EC to the parallelogram CF. Produce BD both ways to the points H, L, and take any number otst;raight Imes BG, GH, each equal to the base BC; andDFC, KL, any number of them, each equal to the base CD ; and join AG, AH, AK, AL : Then, because CB, BG, GH '\ 1. are all equal, the triangles AHG, AGB', ABC are all equal'T Therefore, whatever multiple the base HC is of the base BC, the same multiple is the. triangle AHC of the triangle ABC : For the same reason, whatever multiple the base LC is of the base CD, the same multi- ple is the triangle ALC E A F of the triangle. ADC: And if the base HC be equal to the base CL, the triangle AHC is also equal to the triangle ALO^; and if the base HC be greater than the base CL, likewise the triangle AHC is greater than the triangle ALC; and if,les3-, less : Therefore, since there are four magnitudes, viz. the two bases BC, CD, and the two triangles ABC, ACD ; and of the base BC and the triangle ABC, the first and third, any equi- multiples whatever have been taken, viz. the base HC and tri- angle AHC; and of the base CD and triangle ACD, the se.^ cond and fourth, have been taken any equimultiples whatever, viz. the base CL and triangle ALC ; and that it has been shown, that, if the base HC be greater than the base CL, the triangle AHC is greater than the triangle ALC ; and if equal, equal -^ bA d«f. 5. ^"d if less, less : Therefore^ as the base BC is to the base CD, so is the triangle ABC to the triangle ACD. And because the parallelogram CE is double of the triangle ABC, OF EUCLID. 149 ABO and the parallelogran; CF double of the triangle ACD, ^''^• and that magnitudes have i.he same ratio which their equimul-c 41. 1. tiples hav^e ; as the triangle ABC is to the triangle ACD, so Mj. 5. is the parulicio^ram EC to the parallelogram CF : And because it has been s.io\A'n, that, as the base BC is to the base CD, so is the triangle ABC to t.^e triangle ACD j and as the triangle ABC is to the triangle A_D, so is the parallelogram EC to the para.lelogram CF ; therefore, as the base BC is to tiie base CD, so is^ the parallelogram EC to the parallelogram CF. en.5. Wherefore triangles, &c. Q. E. D. Cor. From this it is plain, that triangles and parallelo- grams that have equal altitudes are one to another as their bases. Let the.r figures be placed so as to have their bases in the same straight line ; and having drawn perpendiculars from the vertices of the triangLs to the bases, the straight line which joins the vertices js parallel to that in which their bases arc*", because the perpendiculars are both equal and pa- '33.1. rallel to one another. Then, if the same construction be made as in the proposition, the demonstration will be the same. PROP. II, THEOR. ^ F a straight line be drawn parallel to one of the see n. sides of a triangle, it shall cut the other sides, or these produced, proportionally : And if the sides, or the sides produced, be cut proportionally, the straight line which joins the points of section shall be parallel to the remaining side of tlie triangle. Let DE be drawn parallel to BC, one of the sides of the triangle ABC : BD is tt^ DA, as CE to EA. Join BE, CD ; then the triangle BDE is equal to the tri- angle CDE'*, because they are on the same base DE, and be- »37. i. twcen the same parallels DE, BC : ADE is another triangle, and equal magnitud s have the same, the same ratio'' j there- "T-S. iu^e^ as the triangle BDE to the triangle ADE,-so is the tri- angle CDE to the triangle ADE , but as the triangleBDE to the triangle ADE, so is<: BD to DA, because having the same c 1. ,-. iltitude, viz. t;.e perpendicular drawn from the point E to AB, ley are to one anoj^er as their basest and for the same reason, L 3 as 150 THE ELEMENTS BooK^. as tiie triangle CDE to the triangle ADE, so is CE to EA : ■uTs, Therefore, as BD to DA, so is CE to EA**. Next, Let the sides AB, AC, of the triangle ABC, or these A. A Ik. D •1.6, f9.5. «3a.i. 3eeN. C B E B C produced, be cut proportionally in the point D, E, that is, so that BD be to DA as CE to EA, and join DE ; DE is paral- lel to BC. The same construction being made, Because as BD to DA, so is CE to EA ; and as BD to DA, so is the triangle BDE to the triangle ADE^ ; and as CE to EA, so is the triangle CDE to the triangle ADE ; therefore the triangle BDE is to the triangle ADE, as the triangle CDE to the triangle ADE ; that is, the triangles BDE, CDE have the same ratio to the triangle ADE ; and thercforc*^ the triangle BDE is equal to the triangle CDE: and they are on the same base DE ; but equal triangles on the same base are between the same paral- lels s ; therefore DE is parallel to BC. Wherefore, if a straight line, &c. 0. E. D. PROP. in. THEOR. xF the angle of a triangle be divided into two eqnril angles, by a straight line M'hich also cuts the bas( , the segments of the base shall have the same ratio whicli the other sides of the triangle liave to one another : And if the segments of tlie base have tlic same ratio which the other sides of the triangle have to one another, the straight line drawn fniin the vertex to the point of section, divides the vci- ticle angle into two equal angles. Let the angle BAC of any triangle ABC be divided into twi equal angles by the straight line AD : BD is to DC, as B/ to AC. Throuul OF EUCLID. 151 ^o.l. Through the point C draw CE parallel to DA, and let BA Book VI. produced meet CE in E. Because the straight line AC meets ^^^^"^^ the parallels AD, EC the angle ACE is equal to the alternate '^ " ' angle CAD*": But CAD, by the hypothesis, is equal to the ^ an^ie BAD; wherefore BAD is equal to the angle ACE. " ' Again, because the straight line BAE meets the parallels AD, K EC, the outward angle BAD is equal to the Inward and ^ opposite angle AEC : but the angle ACE has been proved equal to the angle BAD ; there- fore also ACE is equal to the angle AEC, and consequently iT' the side AE is equal to the side*= AC : And because AD is drawn parallel to one of the sides of the triangle BCE, viz. to EC, BD is to DC, as BA to AE"^, but AE is equal to AC j therefore, as BD, DC, so is <* 2. 6. BAtoAC^ Let now BD be to DC, as BA to AC, and join AD ; the * "' ^' angle BAC is divided into two equal angles by the straight line AD. The same construction being made ; because, as BD to DC, so is BA to AC i and as BD to DC, so is BA to AE"^, because AD is parallel to EC ; therefore BA is to AC, as BA to AE^ : '"• 5. Consequently AC is equal to AEs, and the angle AEC is there- * 9. 5. fore equal to the angle ACE^ : But the angle AEC is equal to " 5.1. the outward and opposite angle BAD ; and the angle ACE is equal to the alternate angle CAD**: Wherefore also the an- gle BAD is equal to the angle CAD : Therefore the angle BAC is cut into two equal angles by the straight line AD, Therefore, if the angle, &c. Q. E. D. L4 152 THE ELEMENTS Book VL, PROP, A. THEOR. If the outward angle of a triangle made by pro-, ducing one of its sides, be divided into two equal angles, by a straight line which also cuts the base produced ; the segments between the dividing line and the extremities of the base have the same ratio which the other sides of the triangle have to one another : And if the segments of the base produced, have the same ratio which the other sides of the tri- angle have, the straight hne drawn from the ver- tex to the point of section divides the outward an- gle of the triangle into two equal angles. Let the outward angle CAE of any triangle ABC bedivided into two 'equal angk's by the straight line AD which meets the base produced in D : BD is to DC, as BA to AC. ?3I. 1. Through CdrawCF parallel to i\D*i and because the straight line AC meets the paiallels AD, tC, the angle ACF is equal "29. 1. to the alternate angle CAD^ ; But CAD is equal to the angle « Hyp. DAE^; therefore' also DAE is equal to the angle ACF. Again, because the straight line FAE meets the parallels AD, FC, the outward angle DAE is equal to the inward and op- j^ posite angle CKA: But the angle ACF has been proved eq^ual to the angle DAE; therefore also the angle ACF is equal to the angle CP'A, and consequently the side AF IS equal to the side «5. 1. ■ AC** : And because AD is parallel to FC, a side of the -tri-. «2. 6. angle BCF, BD is to DC, as BA to AFs but AF is equal to AC ; as therefore BD is to DC, so is BA to AC. Let n )W BD be to DC, as BA to AC, and join AD ; the angle CAD is equal to the angle DAE. T e same construction being made, because BD is to DC, ni. 5. as B.\ to AC ; and that BD is also to DC, as BA, to AF= ; e 9. 5. therefore BA is to AC, as BA to AF*^: wherefore AC is equal b 5^ 1. to AF?, and the angle AFC equal*' to the angle ACF : But the OF EUCLID. 153 the angle AFC is equal to the outward angle EAD, and the ^°°^ ^'^' angle ACF to the alternate angle CAD ; therefore also EAD ^^'^''"*^ is equal to the angle CAD. Wherefore, if the outward, &c. Q.E. D. PROP. IV. THEOR. X HE sides about the equal angles of equiangular triangles are proportionals ; and those which are op- posite to the equal angles are homologous sides, that IS, are the antecedents or consequents of the rajtios. Let ABC, DCE be equiangular triangles, having the angle ABC equal to the angle DCE, and the angle ACB to the an- gle DEC, and consequently^ the angle BAC equal to the angle » 32. 1, CDE. The sides about the equal angles of the triangles ABC, DCE are proportionals ; and those are the homologous sides which are opposite to the equal angles. Let the triangle DCE be placed, so that its side CE may be contiguous to BC, and in the same straight line with it : And because the angles ABC, ACB are together less than two rightangies\ABC,andDEC,which ''^''- ^* is equal to ACB, are also less than F\ two right angles; wherefore BA, I ED produced shall meet<^; let them A \^ 'la. Ax. 1, be produced and meet in the point F ; and because the angle ABC is equal to the angie DCE, BF is pa- rallel'' to CD. Again, because the p. n^^ >. ^,,3 j angle ACB is equal to the angle * DEC, AC is parallel to FE-^: Therefore FACD is a parallelogram ; and consequently AF is equal to CD, and AC to FD'=: And because AC "is parallel «5i. I, to FE, one of the sides of the triangle FBE, BA is to AF, as BC to CE' : But AF is equal to CD ; thereforee, as BA to ^l- 6. CD, so IS BC to CE ; and alternately, as AB to BC, so is DC * '^' ^• to CE : Again, because CD is parallel to BF, as BC to CE, so is FD to DE^ ; but FD is equal to AC; therefore, as BC ' to CE, so is AC to DE •. And alternately, as BC to CA, so CE to DE : rherefore, because it has been proved that AB is to BC, as DC to CE, and as BC to CA, so CE to ED, ex equallS BA is to AC as CD to DE. Therefore the sides, "^ 22- J". &c. Q.E. D. 154 THE ELEMENTS Book Vr, PROP. V. THEOR. ] '23. 1. F the sides of two triangles, about each of their singles, be proportionals, the triangles shall be equi- angular, and have their equal angles opposite to the homologous sides. Let the triangles ABC, DEFhave their sides proportionals, so that A B is to BC, as DE to EF ; and BC to CA, as EF to FD ; and consequently, ex aequali, BA to AC, as ED to DF ; the triangle ABC is equiangular to the triangle DEE, and their equal angles are opposite to the homologous sides, viz. the angle ABC equal to the angle DEE, and BCA to EFD, and also BAG to EDF. At the points E, F, in the straight line EF, makes^ the angle FEG equal to the angle ABC, and the angle EFG equal to BCA ; wherefore the remain- ing angle BAC is equal to the ^' remaining angle EGF'=, and the triangle ABC is there- fore equiangular to the tri- angle GEF ; and consequently they have their -sides opposite to ^Ke- equal angles propor- '*-fi- tionals'=. Wherefore, . as AB to BC, so is GE to EF ; but «' 11.5, as AB to BC, so is DE to EF ; therefore as DE to EF, so*" GE to EF : Therefore DE and GE have the same ratio to EF, and consequently are equal"" : For the same reason DF is, equal to FG: And because, in the triangk^s DEE, GEF, DE is equal to EG, and EF common, the two sides DE, EF, are equal to the two GE, EF, and the base DF is equal to the base GF j therefore the angle DEE is equaK to the angle GEF, and the other angles to the other angles which are subtended by the equal sidess. Wherefore the angle DEE is equal to the angle CFE, and EDF to EGF : And because the angle DEF is equal to the angle GEF, and GEF to the angle ABC ; therefore the angle ABC is equal to the angle DEI" : For the same reason, the angle ACB is equal to the angie DEE, and the angle at A, to tlie angle at D. Therefore the triangle ABC is equiangular to the triangle DEF. Wherefore, if the sides, &c. Q^ E. D. •9. V. :j. e4. 1. OF EUCLID. PROP. VI. THEOR. JL F two triangles have one angle of the one equal to one angle of the other, and the sides about the equal angles proportionals, the triangles shall be equiangular, and shall have those angles equal which are opposite to the' homologous sides. Let the triangles ABG, DEF have the angle BAC in the one equal to the angle EOF in the other, and the sides about those angles proportionals ; that is, B A to AC, as ED to DF ; the triangles ABC, DEF are equiangular, and have the an^Ie ABC equal to the augle DEF, and ACB to DFE. At the points D, F, in the straight line DF, make* the an- '^^c- gl? FDG equal to either of the angles BAC, EDF; and the angle DFG equal to the an- gle ACB : Wherefore the remaining angle at B is equal to the remaining one at G'', and consequently the triangle ABC is equi- angular to the triangle DGF ; and therefore as BA to AC, so is^ GD to DF J but by the hypo- thesis, as BA to AC, so is ED to DF; as therefore ED to DF, so is ^ GD to DF ; wherefore ED is equal '^ to DG; and 11^'^^' Dp' is common to the two triangles EDF, GDF : Therefore the two sides ED, DF are equal to the two sides GD, DF; and the angle EDF is equal to the angle GDF ; wherefore the base EF is equal to the base FG'', and the triangle EDF to "^^ ^• the triangle GDF, and the remaining angles to the remaining angles, each to each, which are subtended by the equal sides : 1 herefore the angle DFG is equal to the angle DFE, and the angle at G to the angle at E : But the angle DFG is equal to the angle ACB ; therefore the angle ACB is equal to the an- gle DFE : A^nd the angle BAC is equal to the angle EDFs ; ' Hj-p. wherefore also the remaining angle at B is equal to the re- maining angle at E. Therefore the triangle ABC is equi- angular to the triangle DEF. Wherefore, if two triangles, kc. Q. E. D. '3*. 1, ♦. G. J56 Book VI. THE ELEMENTS PROP. VII. THEOR. SeeN. »23. I. «'32. 1. ••i. 6. '■=11.5. « 9. 5. ^5. 1. <13. I. iF two triangle ^ have one angle of the one equal to one angle of the other, and the sides ahout two other angles proportionals, then, if each of the re- maining angles be either less, or not less, than a right angle ; or if one of them be a right angle : The triangle shall be equiangular, and have those angles equal about which the sides are proportionals Let the two triangles ABC, DEF have one angle in the one equal to one angle in the other, viz. the angle BAG to the an- gle EDF, and the sides about two other angles ABC, DEF proportionals, so that AB is to BC, as DE to EF ; and, in the iirst case, let each of the remaining angles at C, F be less than a right angle. The triangle ABC is equiangular to the trian- gle OEF, viz. the angle ABC is equal to the angle DEF, and the remaining angle at C to the remaining angle at F. For if the angles ABC, DEF he not equal, one of them is greater than the other : Let ABC be the greater, and at the point B^ in the straight line AB, make the angle ABG equal to the angle ^ DEF ; And because the angle at A is eq^ial to- the angle at D, and the angle ABGr to the angle DEF ; the remaining angle AGB is equal "^ to the remaining angle DFE : Therefore the triangle ABG is equi- angular to t"he triangle DEF ; wherefore '^ as AB is to BG, so is DE to EF ; but as D hi CO EF, se, by hypothesis, is A B to BC ; therefore as AB to BC, so is AB to BG"* : and because AB has the same ratio to each of the lines 'BC, BG ; BC is equal = to BG; and therefore the angle BGC is equal to the an- gle BCG*^ : But the angle BCG is, by hypothesis, less than a right angle ; therefore also the angb BGC is less than a right angle, and the adjacent angle AGB must be greater than a right angle E. But it was proved that the angle GB is equal to the angle at F ; therefore the angle at F is greater than a right angle : But, by the hypothesis, it is less than a right I angle i OF EUCLID. ^51 \T angle ; which is ahsurd. Therefore the angles ABC, DEF l^oo^'Vi. are not unequal, that is, they are equal : And the angls at A 'S ^"^^*^ c;qual to the angle at D j wherefore the remaining angle at C Ts equal to the remaining angle at h : Therefore the triangle ABC is equiangular to the triangle DEF. Next, Let each of the angles at C, F be not less than a right angle : The triangle ABC is also in this case equiangu- lar to the triangle DEF. The same construction being made, it may be j^ proved in like manner that BC is equal to BG, and the angle at C equal to the ^ /A- angle BGC: But the angle t>^ at C is not less than a right C angle ; therefore the angle BGC is not less than a right angle : Wherefore two angles of the triangle BGC, are together not less than two right angles, which is impossible'' ; and therefore the triangle ABC may be " ^'^' ^• proved to be equiangular to the triangle DEF, as in the first case. Lastly, Let one of the angles atC, F, viz. the angle at C, be a right angle; in this case likewise the triangle ABC is equiangular to the triangle DEF. P'or if they be not equian- gular, make, at the point B of the straight line AB, the angle ABG equal to the an- gle DEF ; then it may be proved, as in the first case, that BG is equal to BC : But the angle BCG is a right an- gle,therefore^ the angle BGC is also a right angl.e ; whence two of the angles of the tri- angle BGC are together not less than two right angles, p which isimpessible*": There- fore the triangle ABC is equi- angular to the triangle DEF. Wherefore, if the two triangles, &c, Q. E. D. "^ 15? Book VI. THE ELEMENTS SeeN, n2, 1. "•4. 6. ' 1. clef. 6. PROP. VIII. THEOR. jN a right angled triangle, if a perpendicular be drawn from the right angle to the base ; the trian- gles each side of it are similar to the whole triangle, and to one another. Let ABC be a right angled triangle, having the right angle BAG ; and from the point A let AD be drawn perpendicular to the base BC : The triangles ABD, ADC are similar to the whole triangle ABC, and to one another. Because the angle B AC is equal to the angle ADB, each of them being a right angle, and that the angle at B is common to the two triangles, ABC, ABD : the remaining angle ACB is equal to the remaining angle BAD =^ : Therefore the triangle ABC is equiangular to the triangle ABD, ^nd the sides about their equal angles are proportionals^'' ; wherefore the triangles are similar'^: In • the like manner it may be demonstrated, that the triangle ADC is equiangular and similar to the triangle ABC : And the triangles ABD, ACD, being both equiangular and similar to ABC, are equiangular and similar to each other. There- fore, in a right angled. Sec. Q. E. D. CoR. From this it is manifest that the perpendicular drawn from the right angle of a right angled triangle to the base, is a mean proportional between the segments of the base : And also, that each of the sides is a mean proportional between the base, and its segment adjacent to that side : Because in the triangles BDA, ADC, BD is to DA, as DA to DC' ; and in the' triangles ABC, DBA, BC is to BA, as BA to BD"; and in the triangles ABC, ACD, BC is to C A, as CA to CD". OF EUCLID. Book Vf. PROP. IX. PROB. Jr RO^I a given straight line to cut oiF any partsecN. required. Let AB be the given straight line ; it is required to cutofF any part from it. From the point A draw a straight line AC, making any an- gle with AB ; and in AC take any point D, and take AC the same multiple of AD, that AB is of the part which is to be* cut off from it; join BC, and draw DE parallel to it : Then AE is the part required to be cut off. Because ED is parallel to one of the sides of the triangle ABC, viz. to BC, as CD is to DA, so is * BE to EA : and, by compo- sition^, CA is to AD, as BA to AE : But CA is a multiple of AD; therefore'^ BA is the same multiple of AE : Whatever part therefore AT) is of x\C, AE is the same part of AB: Wherefore, from the straight line AB the part required is cut ofF. .Which was to be done. » 13. 3. PROP. X. PROB. 1 O divide a given straight line similarly to a given divided straight line, that is, into parts that shall have the same ratios to one another which the parts of the divided given straight line have. Let AB be the straight line given to be divided, and AC the divided line : it is required to divide AB similarly to AC. Let AC be divided in the points D, E ; and let AB, AC be placed so as to contain any angle, and join BC, and through the points D, E, draw^ DF, EG parallels to it; and through 131. 1. Ddraw DHK. parallel to AB : Therefore each of the figures FH, HB, is a parallelogram ; wherefore DH is equaPto FG, "3*. 1. and 2.6. i6o T H fe E L E M E N T S ^okVL a„jj HK to GB : And because HE is parallel to KC, one of the sides of the triangle D:vC, as CE to ED, so is "^ KH to HD : But KH is equal to BG, and HD to GF ; therefore, as CE to ED, 'SO is BG to GF : Again, because FD is parallel to EG, one of the sides of the triangle AGE, as ED to DA, so is GF to FA : But it has been proved that CE is to ED, as BG to GF ; and as ilD to DA, so GF to FA : Therefore, the given straight line AB is divided similarly to AC. Which was to be done. PROP. XI. PROB. •2. 6. X O find a tliird proportional to two given straight lines. Let AB, AC be the two given straight lines, and let them be placed so as to contain any angle ; it is required to find a third proportional to AB j AC. Produce AB, AC, to the points D, E ; and make BD equal to AC ; and having joined BC, through D, draw DE parallel to it.» Because BC is parallel to DE, a side of the triangle ADE, AB is '' to BD, as AC ry ^ to CE : But BD is equal to AC ; as there- ^ fore AB to AC, so is AC to CE. Wherefore,' to the two given straight lines AB, AC a third proportional CE is found. Which was to be done. PROP. XII. PROB. 1 O find a fourth proportional to three given straight lines. Let A, B, C be the three given straight lines j it is re- quired to find a fourth proportional to A, B, C. Take OF EUCLID i6i • 31. !. Take two straight lines DE, DF, containing any angle ^^^i EDF ; and upon these make DG equal to A, GE equal to B, and DH equal to C 5 and having joined GH, draw EF parallel » to it through the point E : And because GH is parallel to EF, one of the sides of the triangle DEF, DG is to GE, as DH to HF'' ; but DG is equal to j<^ ft' •> 2. 6. A, GE to B, and DH to C; therefore, as A is to B, so isC to HF. Wherefore to the three given straight lines A, B, C, a fourth proportional HF is found. Which was to be done. PROP. XIII. PROB. JL O find a mean proportional between two given straight lines. Let AB, BC be the two given straight lines ; it is required to find a mean proportional between them. Place AB, BC in a straight line, and upon AC describe the semicircle ADC, and from the point B draw ^ BD at right an- gles to AC, and join AD, DC. Because the angle ADC in a semicircle is a right angle°, and because in the right angled tri- angle ADC, BD is drawn from the right angle perpendicular to the base, DB is a mean propor- tional between AB, BC the segments of the base= : Therefore c cox 8 6. between the two given straight lines AB, BC, a mean propor« tional DB is found. Which was to be done. M » 11. 1. " 31. 3. x62 THE ELEMENTS BookTI. ^ PROP. XIV. THEOR. JiQUAL parallelograms, which have one angle of the one equal to one angle of the other, have their sides about the equal angles reciprocally propor- tional : And parallelograms that have one angle of the one equal to one angle of the other, and their sides about the equal angles reciprocally propor- tional, are equal to one another. Let AB, BC be equal parallelograms, which have the an- gles at B equal, and let the sides DB, BE be placed in the same straight line ^ wherefore also FB, BG are in one straight 'i 14. 1. line*. The sides of the parallelograms AB, BC about the equal angles, are reciprocally proportional j that is, DB is to BE, asGB.toBF. Complete the parallelogram FE ; and because the parallelo- gram AB is equal to BC, and that FE is another parallelo- gram, AB is to FE, as BC to ^rs. p£b. g^t^s ABtoFE, so is ' ^- 6. the base DB to BE^ ; and as BC to FE, so is the base of GB toBF; therefore, as DB to BE, * 21- 5. so is GB to BF'i. Wherefore, the 6ides of the parallelograms AB, BC about their equal an- gles are reciprocally proportional. But, let the sides about the equal angles be reciprocally pro- portional, viz. as DB to BE, so GB to BF; the parellelo- gram AB is equal to the parallelogram BC. Because, as DB to BE, so is GB to BF ; and as DB to BE, so is the parallelogram AB to the parallelogram FE ; and as GB to BF, so is the parallelogram BC to the parallelogram FE i therefore as AB to FE, so BC to FE'' : Wherefore the parallelogram AB is cqual<^ to^he parallelogram BC. There- fore equal parallelograms, &c. Q.ED. « 9. 5. OF EUCLID. PROP. XV. THEOR. il^QUAL triangles which have one angle of the other equal to one angle of the other, have their sides ahout the equal angles reciprocally proportion- al : And triangles which have one angle in the one equal to one angle in the other, and their sides ahout the equal angles reciprocally proportional, are equal to one another. Let ABC, ADE be equal triangles, which have the angle BAG equal to the angle DAE ; the sides about the equal an- gles of the triangles are reciprocally proportiogal j that is, CA is to AD, as EA to AB, Let the triangles be placed so, that their sides, CA, AD be in one straight line ; wherefore also EA and AB are in one straight line* : and join BD. Because the triangle ABCis*!-*. i. equal to the triangle ADE, and that ABD is another triangle ; therefore as the .triangle CAB is to the triangle BAD, so is triangle EAD to triangle DAB** : But as triangle CAB to triangle BAD, ISO is the base CA to AD*^ j and as triangle EAD to triangle DAB, so is the base EA to AB= : as 'therefore CA to AD, so is EA to AB'' ; wherefore the sides of the triangles ABC, ADE about " "• 5. the equal angles are reciprocally proportional. But let the sides of the triangles ABC, ADE about the equal angles be reciprocally proportional, viz. CA to AD, asEA to AB ; the triangle ABC is equal to the triangle ADE. Having joined BD as before j because, as CA to AD, so is EA to AB J and as CA to AD, so is triangle ABC to triangle BAD-; and as EA to AB, so is triangle EAD to triangle ^D' ; therefore'* as triangle BAC to triangle BAD, so is^ ■ :igle EAD to triangle BAD; that is, the triangles BAC, vD have the same ratio to the triangle BAD: Wherefore J triangle ABC is equal' to the triangle ADE. Therefore « 9. 5. -al triangles, &c. 6. E. D. M 2 164 ioot yj. THE ELEMENTS PROP. XVI. THEOR. II. I. 7. 5. 14. 6. XF four straight lines be proportionals, the rectan- gle contained by the extremes is equal to the rectan- gle contained by the means : And if the rectangle , contained by the extremes be equal to the rectangle contained by the means, the four straight lines are proportionals. Let the four straight lines AB, CD, E, F, be proportionals* viz. as AB to CD, so E to F ; the rectangle contained by AB, F is equal to the rectangle contained by CD, E. From the points A, C draw* AG, CH at right angles to AB, CD ; and make AG equal to F, and CH equal to E, and complete theparallellograms BG, DH : Because, as AB to CD, so is E to F ; and that E is equal to CH, and F to AG; AB is'' to CD as CH to AG. Therefore the sides of the paralle- lograms BG, DH about the equal angles are reciprocally pro- portional ; but parallelograms which have their sides about equal angles reciprocally proportional, arc equal to one ano.. ther*^ ; therefore the parallelogram BG is equal to the paralle- logram DH : And the paral- ^ lelogram BG is contained by ^' the straight lines AB, F; be- cause AG is equal to F ; and the parallelogram DH is con- tained by CD and E; because CH is equal to E ; There- fore the rectangle contained by the straight lines AB, F is equal to that which is contained by CD and E. And if the rectangle contained by the straight lines AB, t be equal to that which is contained by CD, E; these four lines are proportional, viz. AB is to CD, as E to F The same construction being made, because the rectang contained by the straight lines AB, F is equal to that which contained by CD, E, and that the rectangle BG is contained ' AB, F, because AG is equal to F ; and the rectangle DH by CD, E, because CH is equal to E ; therefore the parallelo- gram BiG is equal to the parallelogram DH j and they are equi- I OF EUCLID. 165 equiangular : But the sides about the equal angles of equal Boot VI. paralieloff rams-are reciprocally proportional'^ ; Wherefore, as ^"^T'^T^ AB to CD, so is CH to AG ; and CH is equal to E, and AG ' * to F: as therefore AB is to CD, so E to F. Wherefore, if four, &c. Q^E. D. PROP. XVII. THEOR. B- J.F three straight lines be proportionals, therectan- gle contained by the extremes is equal to the square of the mean : and if the rectangle contained by the extremes be equal to the square of the mean, the three straight lines are proportionals. Let the three straight lines A, B, C be proportionals, viz. as A to B, so B to C J the rectangle contained by A, C is equal to the square of B. Take D equal to B j and because as A to B, so B to C, and that B is equal to D • A is * to B, as D to C : But, if four straight lines be pro- portionals, the rectan- \ • gle contained by the extremes is equal to that which is contained by the means'': There- fore the rectangle con- tained by A, C is equal to that contained by B, D : But the rect- angle contained by B, D is the square of B ; because B is equal to D : Therefore the rectangle contained by A, C is equal to the square of B. And if the rectangle contained by A, C be equal to the square of B ; A is to B, as B to C. The same construction being made, because the rectangle contained by A, C is equal to the square of B, and the square of B is equal to the rectangle contained by B, D, because B is equal to D j therefore the rectangle contained by A, C is equal to that contained by B, D ; but if the rectangle con- tained by the extremes be equal to that contained by the means, the four straight lines are proportionals^ : Therefore A is to M 3 B, as 7. #. * 16. 6. i66 O F E U C L I D. Book VI, B, as D to C i but B is equal to D ; wherefore, as A to B, '"^'"'''^^ so B to C i Therefore, if three straight lines, Sec. Q. E. D. SeeN. PROP. XVIII. PROB. U PON a given straight line to describe a rectili- neal figure similar, and similarly situated to a given rectilineal figure a" Let AB be the given straight line, and CDEF the given rectilineal figure of four sides; it is required upon the given straight line AB to describe a rectilineal figure similar, ^nd similarly situated to CDEF. Join DF, and at the points A, B in the straight line AB, ■> :.5. 1. malce^ the angle BAG equal to the angle! at C, and the angle ABG equal to the angle CDF ; therefore the remaining angle ■•'*" • CFD is equal to the remaining angle AGB^ Wherefore the triangle FCD is equiangular, to the jr l\ triangle GAB: (^ Again, at the points G, B in the straight line GB, make * tiie angle BGH equal to the angle DFE, and the angle G B H, A equal to F D E ; therefore the remaining angle FED Is equal to the remaining angle GHB, and the triangle FDE equiangular to the triangle GBH : Then, because the angle AGB is equal to the angle CFD, and BGH to DFE, the whole angle AGH is equal to the whole CFE: For the same reason, the angle ABH is equal to the angle CDE ; also the angle at A is equal to the angle at C, and the angle GHB to FED : Therefore the rec- tihneal figure ABHG is equiangular to CDEF : But likewise these figures have their sides about the equal angles propor- tionals ; because the triangles GAB, FCD being equiangular, .. ■;. BA is <= to AG, as DC to CF ; and because AG is to GB, as CF to FD ; and as GB to GH, so, b}' reason of the equiangu- ■^2. :j. lar triangles BGH, UFE, is FD to FE; therefore, ex aequali'', AG is to GH, as CF to FE : In the same manner it may be proved that AB is to BH, as CD to DE : And GH is to HB, as OF EUCLID. 167 HB, as FE to ED=. Wherefore, because the rectilineal figures Book vi. ABHG, CDEF are equiangular, and have theirsidesabout the ^•-^•'''*^ equal angles proportionals, they are similar to one anothor*. « i_ jigf. g. Next, let it be required to describe upon a given straight line AB, a rectilineal figure similar, and similarly situated to the rectilineal figure CDKEF. Join DE, and upon the given straight line AB describe the rectilineal figure ABHG similar, and similarly situated to the quadrilateral figure CDEF, by the former case j and at the points B, H, in the straight line BH, make the angle HBL etjual to the angle EDK, and the angle BHLequal to the angle DEK ; therefore the remaining angle at K is equal to the re- mainin;^ angle at L : And because the figures ABHG, CDEF are similar, the angle GHB is equal to the angle FED, and BHL is equal to DEK ; wherefore the w^hole angle GHL is equal to the whole angle FEK: For the same reason the angle ABL is equal to the angle CDK : Therefore the five -sided fio'ures AGHLB, CFEKD are equiangular ; and because the figures AGHB, CFED are similar, GH is to HB, as FE to ED ; and as HB to HL, so is ED to EK= ; therefore, ex sequa- li-, GH is to HL, as FE to EK : For the same reason, AB is ' *■ «• to BL as CD to DK : And BL is to LH, as-^ DK to KE, be- « 22. 5. causethe triangles BLH, DKF are equiangular : Therefore, because the five-sided figures AGHLB, CFEKD are equian- gular, and have their sides about the equal angles proportion- als, they are similar to one another ; and in the same manner a rectilineal figure of six or more sides may be described upon a given straight line similar to one given, and so on. Which was to be done. PROP. XIX. THEOR. Similar triangles are to one another in the du- plicate ratio of their homologous sides. Let ABC, DEF be similar triangles, having the angle B . equal to the angle E, and let AB be to BC, as DE to EF, so that the side BC is homologous to EF^ : the triangle ABC * ^2. def. has-fio the triangle DEF, the duplicate ratio of that which BC has to EF. Take BG a third proportional to BC, EF*': so that BC is*" 11. 6. to EF, as EF to BG, and join GA : Then, because as.AB to BC, so DE to EF j alternately^ AB is to DE, as BC to ' ic 5. M4 EF; i68 THEELEMENTS ^K VI. EF : But as BC to EF, so is EF to BG ; thereforeJ as AB o II ^^ to DE, so is EF to BG : Wherefore the sides of the triangles ABG, DEF, which are about the equal angles, are recipro- cally proportional : But triangles which have the sides about two equal angles re- ciprocally proportio- j^ - nal are equal to one * 15. 6. another'^ : I'herefore the triangle ABG is equal to the triangle DEF: And because as BC is to EF, so EF to BG ; anh that if three straight lines « '10. def.5. be proportionals, the first is said *' to have to th6 third the du- plicate ratio of that which it has to the second ; BC therefore has toBG the duplicate ratio of that which BC has to EF : e I.e. ' But as BC to BG, so iss the triangle ABC to the triangle ABG. Therefore the triangle ABC has to the triangle ABG the duplicate ratio of that which BC has to EF : But the tri- angle ABG is equal to the triangle DEF : Wherefore also the triangle ABC has to the triangle DEF the duplicate ratio of that which BC has to EF. Tiierefore similar triangles, &c. Q, E. D. . Cor. From this it is manifest, that if three straight lines be proportionals, as the first is to the third, so is any triangle upon the first to a similar, and 'similarly described triangle upon the second. PROP. XX. THEOR. oLMILAR polygons may be divided into thesame number of similar triangles, having the same latio to one another that the polygons have; and the poly- gons have to one another the duplicate ratio of that which their liomoloo-ous sides have. Let ABCDE, FGHKL be similar polygons, and let AB be the honiologous side to FG: The polygons ABCDE, FGHKL may be divided into the same number of similar tri- angles, whereof each to each has the same ratio which the po- lygons have ; and the polygon ABCDE has to the polygon FGHKL the duplicate ratio of that which the side AB has to the side FG. Join BE, EC;GL, LH : And because the polygon ABCDE is OF EUCLID. 169 is similar to thepolygon FGHKL, the angle BAE is equal to the ^'^• angle GFL*, and BA is to AE, as GF to FL=^: Wherefore, , ^ ^^j ^^ because the triangles ABE, FGL have an angle in one equal to an an^^le in the other, and their sides about these equal angles proportionals, the triangle ABE is equiangular'', and there- " 6. 6. fore similar to the triangle FGL<^ ; wherefore the angle ABE "i-S. ^ is equal to the angle FGL : And, because the polygons are simi- lar, the whole angle ABC is equalHo the wliole angle FGH ; therefore the remaining angle EBC is equal to the remaining angle LGii : And because the triangles ABE, FGL are similar, EB is to BA, as LG to GF»; and also, because the polygons are similar, AB is to BC, as EG to GH^ therefore, ex aequali'^, * 22. 5. EB is to BC, as LG to GH ; that is, the sides about the equal aneles EBC, LGH are proportionals; therefore'^ the tnano-le EBC is equiangular to the triangle LGH, and similar to its For the same rea- ^ son, the tri- ^,x^'^^s. M ^ angle ECD ^^ ^^^ X, r >> likewise is '^;^~ y ^ ^ similar to V^\ / TJw /" the triangle LHK: there- fore the si- milar poly- gons ABCDE, FGHKL are divided into the same number of similar triangles. Also these triangles have, each to each, the same ratio which the polygons have to one another, the antecedents being ABE, EBC, ECD, and the consequents FGL, LGH, LHK : And the polygon ABCDE has to the polygon FGHKL the dupli- cate ratio of that which the side AB has to the homologous side FG. Because the triangle ABE is similar to the triangle FGL, ABE has to FGL, the duplicate ratio<^ of that which the side « 19. o BE has to the side GL : For the same reason, the triangle BEC has to GLH the duplicate ratio of that which BE has to GL : Therefore, as the triangle ABE to the triangle FGL, so*^ is the^ n. ,. triangle BEC to the triangle GLH. Again, because, the tri- angle EBC is similar to the triangle LGH, EBC has to LGH, the duplicate ratio of that which the side EC has to the side LH : For the same reason, the triangle ECD has to the triangle LHK, lyo THE ELEMENTS Book VI. LHK, the duplicate ratio of that which EC has to LH : As >^r*^ therefore the triangle EBC to the triangle LGH, so is*" the '11. 5. . , ^ ^ triangle ECD to the A triangle ^^-'•^^\ M LHK: But it has been proved, that the triangle EBC is like- wise to the triangle LGH, as the triangle ABE to the triangle FGL. Therefore, as the triangle ABE to the triangle FGL, so is triangle EBC to triangle LGH, and triangle PXD to triangle LHK : And therefore, as one of the antecedents to one of the consequents, s 12. 5. so are all the antecedents to all the consequents s. Wherefore, as the triangle ABE to the triangle FGL, so is the polygon ABCDE to the polygon FGHKL : But the triangle ABE has to the triangle FGL, the duplicate ratio of that which the side AB has to the homologous side FG. Therefore also the poly- gon ABCDE has to the polygon FGHKL the duplicate ratio of that which AB has to the homologous side EG. Where- fore similar polygons, &c. Q^ E. D. Cor. I. in like manner, it may be proved, that similar four sided figures, or of any number of sides, are one to another in the duplicate ratio of their homologous sides, and it has already been proved in triangles. Therctore, universally, similar rec- tilineal figures are to one another in the duplicate ratio of their homologous sides. Cor. 2". And if to AB, FG, two of the homologous sides, h lO.dcf.5. a third proportional M be taken, AB has** to M the duplicate ratio of that which AB has to FG : but the four-sided figure or polygon upon AB, has to the four-sided figure or polygon upon FG likewise the duplicate ratio of that which AJ5 has to FG : Therefore, as AB is to M, so is the figure upon AB to the *Cor.j9.6. %'J''s upon FG, which was also proved in triangles'. Therefore, universally, it is manifest, that if three straight lines be proportionals, as the third is to the third, so is any rec- tilineal figure uptfn the first, tb a similar and similarly de- scribedrectilineal figure upon the second. OF EUCLlb. PROP, XXI. THEOR. X\ E c T I L i^N ZA L figuics wliich are similar to the same rectilineal figure, are also similar to one another. Let each of the rectilineal figures A, B be similar to the rec- tilineal figure C : The figure A is similar to the figure B. Because A is similar to C, they are equiangular, and also have their sides about the equal angles proportionals'. Again, » i. d*Li. because B is similar to C, they are equian- gular, and have their sides about the equal angles proportionals*. Therefore the figures A, B are each of them equiangular to C, and have the sides about the equal angles of each of them and of B proportionals. Wherefore the rectili- neal figures A and C are equiangular'', and have their sides " i. Ax. i. about the equal angles proportionals'". Therefore A is similar* "^ J'- J- to B. Q, E. D. ^ PROP. XXII. THEOR. J-F four straight lines be proportionals, the similar rectilineal figures similarly described upon them shall also be proportionals ; and it' the similar rec- tilineal figures similarly described upon four straight lines be proportionals, those straight lines shall be proportionals. Let the four straight lines AB, CD, EF, GH be propor- tionals, viz. AB to CD, as EF to GH, and upon AB, CD let the similar rectilineal figures KAB, LCD be similarly de- scribed ; and upon EF, GH the similar rectilineal figures VIF, NH, in like manner : The rectilineal figure KAB is to XD, as MF to NH. To AB, CD take a third proportional* X; and to EF, GH » ii. «. I third proportional O : And because AB, is to CD, as EF to 3H, and that CD is'' to X, as GH to O; wherefore, ex Ml. 5. qiuliS as AB to X, so EF to O : But as AB to X, so is the c oo. 5. rectiliaeal 172 THE ELEMENTS £ooK VI. rectilineal KAB tothfe rectilineal LCD, and as EF to O, so is ^ the rectilineal MF to the rectilineal NH : Therefore, as KAB to LCD, so" is MF to N H. And if the reailineal KAB be to LCD, as MF to NH'; the straight line AB is to CD, as EF to GH. «12. 6. Malce'= as AB to CD, so EF to PR, and upon PR describe' the rectilineal figure SR similar and similarly situated to either X »"18. 6. E F <i H P K of the figures MF, NH : Then, because as AB to CD, so is EF to PR, and that upon AB, CD are described the similar and similarly situated rectilineals KAB, LCD, and upon EF, PR, in like manner, the similar rectilineals MF, SR; KAB is to LCD, as MF to SR; but by the hypothesis KAB is to LCD, as MF to NH ; and therefore the rectilineal MF having the t. 9, 5. same ratio to each of the two NH, SR, these are equal g to one another : They are also similar, and similarly situated ; there- fore GH is equal to PR : And because as AB to CD, so is EP to PR, and that PR is equal to GH ; AB is to CD, as EF tc GH. If therefore four straight lines, &c. Q. E. D. PROP. XXin. THEOR. SeeN. liQuiANGyLAR parallelogTams have to one ano- ther the ratio which is compounded of the ratios oS their sides. j Let AC, CF be equiangular parallelograms, having the angll BCD equal to the angle ECG: The ratio of the parallelogi am AC to the parallelogram CF,is the same with the ratio which is compounded of the ratios of their sides. L^ I OF EUCLID. 173 • Let BC, CG be placed in a straight line; therefore DC and Book vi, CE are also in a straight line*; and complete the parallelogram ^^^ DG ; and talcing any straight line K, make*' as BC to CG, " 12! 6. so K to L ; and as DC to CE, so make^ L to M : Therefore, the ratios of K to L ; and L to M, are the same with the ratios of the sides, viz, of BC to CG, and DC to CE. But the ratio of K to M is that which is said to he compounded '^ of the « A. dcf. j. ratios of K to L, and L to M : Wherefore also K has to M the ratio compounded of the ratios of the sides : And because as BC to CG, so is the parallelogram AC to the parallelogram CH' ; but \ \ \ • «ii. c. as BC to CG, so is K to L ; there- fore K is ^ to L, as the parallel- 1 \ \ «ii.5. ogram AC to the parallelogram CH : Again, because as DC to CE, so is the parallelogram CH to the parallelogram CF ; but as DC to CE, so is L to M ; where- fore L is f to M, as the paral- lelogram CH to the parallelogram CF : Therefore since it has been proved, that as K to L, so is the parallelogram AC to the parallelogram CH; and as L to M, so the parallelogram CH to tr.e parallelogram CF ; ex aequali*^, K is to M, as the pa- f 33. 5, rallelogram AC to the parallelogram CF : But K has to M the rat'io which is compounded of the ratios of the sides; therefore also the parallelogram AC has to the parallelogram CF the ratio which is compounded of the ratios of the sides. Where- ' fore ecjuiangular parallelograms, &c. Q. E. D. PROP. XXIV. THEOR. i HE parallelograms about the diameter of any sceN. parallelogram, are similar to the whole, and tp one another. Let ABCD be a parallelogram, of which the diameter is AC ; and EG, HK the parallelograms about the diameter : The parallelograms EG, HK are similar both to the whole parallelogram ABCD, and to one another. Because DC, GF are parallels, the angle ADC is equal* to » 2?. 1. •he angle AGF : For the same reason, because BC, EF are pa- rallels, 174 *. 6. «1. clef. 6. n-.e. SeeN. "Cor. 45.1 129, 129, 1. 1, « 13. 6. ■» 18. 1. ' 2. Cor. 2(). 6. THE ELEMENTS rallels, the angle ABC is equal to the angle AEF : And eacK of the angles BCD, EFG is equal to the opposite angle DAB*", and therefore are equal to one another, wherefore the paral- lelograms ABCD, AEFG are equiangular : And because the angle ABC is equal to the angle AEF, and the angle BAG common to the two triangles BAC, EAP\ they are equiangu- lar to one another ; therefore*^ as AB toBC, sois AEtoEF: And because the opposite sides of parallelograms are equal to one another^, AB'' is to AD, as AE to AG ; and DC to CB as GF to FE ; and also CD to DA, as FG to GA ; Therefore the sides of the parallelograms ABCD, AEFG about the equal angles are proportionals ;' and they are there- fore similar to one another^ : For the same reason the paral- lelogram ABCD is similar to the parallelogram FHCK. Wherefore each of the parallelograms GE, KH is similar to DB : But reftilineal figures which are similar to the same rectilineal figure are also similar to one another*^ ; therefore the parallelogram GE is similar to KH. Wherefore the pa. rallelograms, &c. Q^ E. D. 'PROP. XXV. PROB. 1 O describe a rectilineal figure which sliall be si- milar to one, and equal to another given rectilineal figure. Let ABC be the given rectilineal figure, to which the figure to be described is required to be similar, and D that to which it must be equal. It is required to describe a rectilineal figure similar to ABC, and equal to D. Upon the straight line BC describe'' the parallelogram BE equal to the figure ABC; also upon CE describe '^ the paral- lelogram CM equal to D, and having the angle FCE equal to the angle CBL: Therefore BC and CF are in a straight line^, as also LE and EM : Between BC and CF find"^ a mean proportional GH, and upon GH describe'' the redilineal figure KGH similar and similarly situated to the figure ABC: And because BC is to GH as GH to CF, and if three straight lines be proportionals, as the first is to the third, so is '^ the %ure OF EUCLID. figure upon the first to the similar and similarly described fi- ^°ok vi. gure upon the second ; therefore as BC to CF, so is the rec- ^^^ tilineal figure ABC to KGH : But as BC to CF, so is*^ the pa- *" i. 6. rallelogram BE to the parallelogram ^F : Therefore as the rectilineal figure ABC is to KGH, so is the parallelogram BE to the parallelogram EFs ; and the rectilineal figure ABC is« ji. 5. A equal to the parallelogram BE ; therefore the rectilineal figure KGH is equal'' to the parallelogram EF : But EF is equal' to the figure D ; wherefore also KGH is equal to D ; and it is similar to ABC Therefore the rectilineal figure KGH has been described similar to the figure ABC, and equal to D. Which was to be done. u, 5, PROP. XXVI. THEOR. IF twosimilar parallelograms have a common an- gle, and be ;^imilarly situated ; they are about the same diameter. Let the parallelograms ABCD, AEFG be similar and simi- larly situated, and have the angle DAB common. ABCD and AEFG are about the same diameter. For, if not, let, if possible, the parallelogram BD have its dia- j meter AHC in a different straight ! line from AF, the diameter of the I parallelogram, EG, and let GF meet AHC in H; and throuc^h jH draw HK parallel to AD or { BC : Therefore the parallelograms I ABCD, /\KHG being about the same diameter, they are similar to ©ne another^ : Wherefore as DA to AB, so is*^ GA to AK : 2 But • 24. «. » l.-def. C. 176 THE ELEMENTS i;ooK VI. But bqcause ABCD and AEFG arc similar parallelograms, ^"^^^^^ as DA is to AB, so is GA to AE j therefore'^ as GA to AE, so GA to AK J wherefore GA has the same ratio to each of * 9. 5. the straight lines AE, AK ; and consequently AK is equal^ to AE, the less to the greater, which is impossible : Therefore ABCD and AKHG are not about the same diameter ; where- , fore ABCD and AEFG must be about the same diameter. Therefore if two similar, &c. Q^ E. D. ' To understand the three following propositions moreea- * sily, it is to be observed, * 1. That a parallelogram is said to be applied to a straight ' line, when it is described upon it as one of its sides. Ex. gr. ' the parallelogram AC is said to be applied to the straight * line AB. ' 2. But a parallelogram AE Is said to be applied to a straight * line AB, deficient by a parallelogram j when AD the base of * AE is less than AB, and there- ' * fore AE is less than the paral- ' lelogram AC described upon * AB in the same angle, and ' between the same parallels, by ' the parallelogram DC; and ' DC is therefore called the 'defect of AE. * 3. And a parallelogram AG is said to be applied to a ' straight line AB, exceeding by a parallelogram, when" AF ' the base of AG is greater than AB, and therefore AG ex- ' ceeds AC the parallelogram described upon AB in the same ' angle, and between the same parallels, by the parallelogram * BG.' 5CC N, E ( c G " A. i) J s V PROP. XXVII. THEOR. v^F all parallelograms applied to the same straight line, and deficient by parallelograms, simi- lar and similarly situated to thatwliichis described upon the half of the line ; that which is applied to the half, and is similar to its defect, is the greatest. Let AB be a straight line divided into two equal parts inC, and let the parallelogram AD be applied to the half AC, which is therefore deficient from the parallelogram upon the whole line AB by the parallelogram CE upon the other half CB: OF EUCLID. CB : Of all the parallelograms applied to any other parts of AB, and deficient by parallelograms that are similar, and simi- larly situated to CE, AD is the greatest. Let AF be any parellelogram applied to AK, any other part of AB than the half, so as to be deficient from the parallel- ogram upon the whole line AB by the parallelogram KH si- mitar and similarly situated toCE : AD is greater than AF, First, let AK the base of AF, be greater than AC the half of AB ; and because CE is similar to the parallelogram KH,they are about the same diameter* : Draw their diameter DB, and complete the scheme: Be- cause the parallelogram CF is equal*" to FE, and KH to both, therefore the whole CH is equal to the whole KE : But CH is equa^ to CG, because the base AC is equal to the base CB; therefore CG is equal to KE : To each of these add CF i then the whole AF is equal to the gnomon CHL ; Therefore CE, or the parallelogram AD, is greater than the parallelogram AF. Next, let AK, the base of AF, be less than AC, and, the same construc- tion being made, the parallelogram DH is equal to DGs for HM is equal to MG'', because BC is equal to CA ; wherefore DH is greater thanLG: But DH is equalno DK j therefore DK is greater than LG : To each of these add AL ; then the ■whole AD is greater than the whole AF. Therefore, of all parallelograms applied, &c. Q. E. D. J77 Book VI. ^36. I. KB *^. 1. N 178 Book VI. Sec N. » 10. 1. 18. 6. •< 21. 6. THE ELEMENTS PROP. XXVIII. PROB. J. O a given straight line to apply a parallelogram equal to a given rectilineal figure, and deficient by a parallelogram similar to a given parellelogram : But the given rectilineal figure to which the paral- lelogram to be applied is to be equal, must not be greater than the parallelogram applied to half of the given line, having its defect similar to the defect of that which is to be applied ; that is, to the given parallelogram. Let AB be the'given straight line, and C the given rectilineal figure, to which the parallelogram to be applied is required to be equal, which figure must not be greater than the paral- lelogram applied to the half of the line having its defect from that upon the whole line similar to the defect of that which is to be applied ; and let D be the parallelogram to which this defect is required to be similar. It is required to apply a pa- rallelogram to the straight line AB, which shalfbe H G_ OF equal to the figure C, and be deficient from the paral- lelogram upon the wioJe line by a parallelogram si- milar to D. Divide AB into two equal f arts=» in the point E, and upon EB describe the pa- rallelogram EBFG similar'' and similarly situated to D, and complete the parallelo- gram AG, which must cither be equal to C, or greater than it, by the determination : And if AG be squal to C, then what was required is already dond : For, upon the straight line AB, the parallelogram AG is applied equal to the figure C, and deficient by the parallelo- gram EF similar to D : But, if AG be not equal to C, it is greater than it ; and EF is equal to AG ; therefore EF also is greater than C. Make c the parallelogram KLMN equal to" the excess of EF above C, and similar and similarly situated to D i but D is similar to EF, therefore* also KM is similar I to OF EUCLID. 179 to EF : Let KL be the homologous side to EG, and LM to Book VL GF : and because EF is equal to C and KM together, EF ^^'^''^^ is greater than KM ; therefore the straight line EG is greater than KL, and GF than LM : Make GX equal to LK, and GO equal to LM, and complete the parallelogram XGOP : Therefore XO is equal and similar to KM ; but KM is simi- lar toEF ; wherefore also XO is similar to EF, and therefore XO and EF are about the same diameter* : LetGPB be their * -6. o. diameter, and complete the scheme : Then because EF is equal to C and KM together, and XO a part of the one is equal to KAI a part of the other, the remainder, viz. the gno- mon ERO, is equal to the remainder C : and because OR is equal ^ to XS, by adding SR to each, the whole CB is equal f 34 2. to the whole XB : But XB is equals to TE, because the base g 3^ ^ AE is equal to the base EB j wherefore also TE is equal to OB; Add XS to each, then the whole TS is equal to the whole, viz, to the gnomon ERO : But it has been proved that the gnomon ERO is equal to C, and therefore also TS is equal to C. Wherefore the parallelogram TS, equal to the given rectilineal figure C, is applied to the given straight line AB deficient by the parallelogram SR, similar to the given one D, because SR is similar to EF**. Which was to be done. " 24. «, PROP. XXIX. PROB. JL O a given straight line to apply a parallelogram seeN. equal to a given rectilineal figure, exceeding by a parallelogram similar to another given. Let AB be the given straight line, and C the given rectili- neal figure to which the«parallelogram to be applied is required to be equal, and D the parallelogram to which the excess of the one to be applied above that upon the given line is required to be similar. It is required to apply a parallelogram to the given straight line AB which shall be equal to the figure C, exceeding by a parallelogram similar to D. Divide AB into two equal parts in the point E, and upon » 13. e, EB describe ' the parallelogram EL similar, and similarly situ- N 2 ated i8o THE ELEMENTS Book VI. ated to-D : And make ^ the parallejogram GH equal t6 EL and C together, and similar, and similarly situated to D ; where- fore CH is similar to EL*^ : Let KH be the side homologous / to FL, and KG to FE : And because the parallelogram GH is greater than EL, therefore the side KH is greater than FL, and KG than FE : Produce FL and FE, and make FLM equal to KH,and FEN toKG,and complete the parallelogram MN. MN is therefore equal and similar to GH j . ' but GH is similar to EL ; wherefore MN is similar to EL, and consequently EL and MN are about the "26.6. same diameter^: Draw their diameter FX, and complete the scheme. Therefore, .since GH is equal to EL and C together, and that GH is equal toMN J MN is equal to EL and C : Take away the common part EL ; then the remainder, via. the gnomon NOL, is equal to C. And be- cause AE is equal to EB, the parallelogram AN is equal= to the parallelogram NB, that is, to BM^. Add NO to each ; therefore the whole, viz. the parallelogram AX, is equal to the gnomon, NOL^ But the gnomon NOL is equal to C ; there- fore also AX is equal to C. Wherefore to the straight line AB there is applied the parallelogram AX equal to the given rectilineal C, exceeding by the parallelogram PO, which i& • 24. 6. similar to D, because PO is similar to ELs. Which was t* be done. • 36. 1. '«. 1. PROP. XXX. PROB. To ratio. cut a given straight line in extreme and mean Let AB be the given straight linej it is required to cut it in extreme and mean ratio. 1 Upon B OF EUCLID. Upon AB describe » the square BC, and to AC apply the parallelogram CD equal to BC, exceeding by the %ure AD similar to BC'': But BC is a square, therefore also AD is a square j and be- cause BC is equal to CD, by taking the common part CE from each, the re- mainder BF is equal to the remainder J^ AD : And these figures are equiangular, therefore their sides about the equal an- gles are reciprocally proportional c : Wherefore, as ¥E to ED, so AE to EB : But FE is equal to AC\ that is, to AB ; and ED is equal to AE : Therefore as BA to AE, so is AE to EB : But AB ^ is greater than AE ; wherefore AE is greater than EB= : Therefore the straight line AB is cut in extreme and mean ratio in E'. Which was to be done. Otherwise, Let AB be the given straight line ; it is required to cut it in extreme and mean ratio. Divide AB in the point C, so that the rectangle contained by AB, BC be equal to the square of AC s : Then, because the rectangle AB, BC is -y: p p equal to the square of AC, as BA to AC, so is AC to CB ^ : Therefore AB is cut in extreme and mean ratio in C*^. Which was to be done. iSi Book VI. E ■ ' ' 14. 6, '^ .34. 1. « H. 5. '^ J def. 6. t tl. Ot » n. 6. PROP. XXXL THEOR. IN right angled triangles, the rectilineal figure de- See^u jcribed upon the side opposite to the right angle, is ?qual to the similar, and similarly described figures ipon the sides containing the right angle. Let ABC be a right angled triangle, having the right angle 5AC : The rectilineal figure described upon BC is equal to le similar, and similarly described figures upon BA, AC' Draw the perpendicular AD ; therefore, because in the right igled triangle ABC, AD is drawn from the right angle at A erpendicular to the base BC, the triangles. ABD, ADC are limilar to the whole triangle ABC, and to one another^ and i g. g. N 3 because 182 ' 2 Cor. 00. 6. THE ELEMENTS because the triangle ABC is similar to ADB, as CB to BA, s© is BA to BD '' ; and because these three straight lines are pro- portionals, as the first to the third, so is the figure upon the first to the similar, and similarly described figure upon the second^ : Therefore as CB to BD, so 'is the figure CB to the similar and larly described figure upon simi- upon * B. 5. BA': And inversely'', as DB to BC, so is the figure upon BA to that upon BC : For the same reason, as DC to CB, so is the figure upon CA to that upon CB. Wherefore as BD and DC t9gether to BC, so are the figures upon BA, e 2i. 5. AC to that upon BC " : But BD and DC together are equal f A. 5. to BC. Therefore the figure described on BC is equaK to the similar and similarly described figures on BA, AC. Where- fore, in right angled triangles, &c. Q^. E. D. PROP. XXXII. THEOR. SeeN. JLy two tHanglcs which have two sides of the one proportional to two sides of the other, be joined at one angle, so as to have their homologous sides pa- rallel to one another ; the remainins: sides shall be in a straight line. Let ABC, DCE, be two triangles which have the two sides: | BA, AC proportional to the two CD, DE, viz. BA to AC, as CD to DE ; and let AB be parallel to DC, and AC to DE,- BC and CE are in a straight line. Because AB is parallel to \ DC, and the straight line AC meets them, the alter- nate angles BAC, ACD * 29. 1, sre equal* ; for the same reason, the angle CDE is equal to the angle ACD ; wherefore also BAC is equal to CDE : And because OF EUCLID. 183 the triangles ABC, DCE have one angle at A equal to one at ^poR vi. D, and the sides about these angles proportionals, viz. BA to ^''^^''^'^■ AC, as CD to DE, the triangle ABC is equiangular'' to DCE : " 6. 6. Therefore the angle ABC is equal to the angle DCE : And the angle BAC was proved to be equal to ACD : Therefore the whole angle ACE is equal to the two angles ABC, BAC ; add the common angle ACB, then the angles ACE, ACB are equal to the angles ABC, BAC, ACB : But ABC, BAC, ACB are equal to two right angles'^ ; therefore also the angles « 352. 1. ACE, ACB are equal to two right angles : And since at the point C, in the straight line AC, the two straight lines BC, CE, which are on the opposite sides of it, make the adjacent angles ACE, ACB equal to two right angles ; therefore"* BC " i-^- J- and CE are in a straight line. Wherefore, if two triangles, &c. Q. E. D. PROP. XXXIII. THEOR. J. N equal circles, angles, whether the centres' or seeN. circumferences, have the same ratio which the cir- ' cumferenccs on which they stand have to one ano- ther : So also have the sectors. Let ABC, DEF be equal circles ; and at their centres the angles BGC, EHF, and the angles BAC, EDF, at their cir- cumferences J as the circumference BC to the circumference EF, so is the angle BGC to the angle EHF, and the angle BAC to the angle EDF ; and also the seclor BGC to the sector EHF. Taiceany number of circumferences CK, KL, each equal to BC, and any number whatever FM, MN, each equal to EF : "And join GK, GL, HM, HN. Because the circumferences BC, CK, KL are all equal, the angles BGC, CGK, KGL are also all equal*: Therefore what multiple soever the circum^ * 27.3. ference BL is of the circumference BC, the same multiple is the angle BGL of the angle BGC : For the same reason, what- ever multiple the circumference EN is of the circumference EF, the same multiple is the angle EHN of the angle EHF ; N4 And i84 THE EI^EMENTS Book VI. And if the circumference BL be equal to the circumference .^"^iPg"**^ EN, the angle BGL is also equal'' to the angle EHN ; and if the circumference BL be greater than EN, likewise the angle BGL is greater than EHN ; and if less, less : There being then four magnitudes, the two circumferences BC, EF, and the two angles BGC, EHF ; of the circumference BC, and of the angle BGC, have been taken any equimultiples whatever, viz. the circumference BL, and the angle BGL ; and of the circum- ference EF, and of the angle EHF, any equimultiples what- ever, viz. the circumference EN, and the angle EHN : And it has been proved, that, if the circumference BL, be greater than EN, the angle BGL is greater than EHN j and if equal, equal; and if less, less: As therefore the circumference BC •• 5. def. 5. to the circumference EF, so ^ is the angle BGC to the angle EHF : But as the angle BGC is to the angle EHF, so is « 13. 5. *" the angle BAC to the angle EDF ; for each is double of <20. 3. cach^; Therefore, as thel circumference BC is to EF, so is the angle BGC to the angle EHF, and the angle BAC to the angle EDF. Also, as the circumference BC to EF, so is the sedlor BGC to the senior EHF. Join BC, CK, and in the circumferences BC,CKtake any points X, O, and join BX, XC, CO, OK: Then, because in the triangles GBC, GCK the two sides BG^ GC are equal to the two CG, GK, and that they contain • 4. 1. equal angles ; the base BC is equal = to the base CK, and the triangle GBC to the triangle GCK : And because the circum- ference BC is equal to the circumference CK, the remaining part of the whole circumference of the circle ABC, is equal to the remaining part of the whole circumference of the same circle : Wherefore the angle BXC is equal to the angle COK»; 'n.^def.3. and the segment BXC is therefore similar to the segment COK*^; and OF EUCLID. iSs and they are upon equal straight lines BC, CK : But similar ^°°^ '^'• segments of circles upon equal straight lines, are equals to one ^H^^'^ another: Therefore the segment BXC is equal to the segment COK : And the triangle BGC is equal to the triangle CGKj therefore the whole, the sector BGC, is equal to the whole, the SecSlor CGK : For the same reason, the senior KGL is equal to each of the seftors BGC, CGK : In the same manner, the sedors EHF, FHM, MHN may be proved equal to one ano- ther : Therefore, what multiple soever the circumference BL is of the circumference BC, the same multiple is the sector BGL of the se£lor BGC : For the same reason, whatever multiple the circumference EN is of EF, the same multiple is the seftor EHN of the sedor EHF : And if the circumference BL be B-^: equal to EN, the sector BGL is equal to the sedor EHN ; and if the circumference BL be greater than EN, the sedtor BGL is greater than the sedor EHN; and if less, less: Since then, there are four magnitudes, the two circum- ferences BC, EF, and the two sedors BGC, EHF, and of the circumference BC, and sedor BGC, the circumference BL and sedor BGL are any equalmultiples whatever; and of .the circumference EF, and sedor EHF, the circumference EN, and sedor EHN, are any equimultiples whatever ; and that it has been proved, if the circumference BL be greater than EN, the sedor BGL is greater than the sedor EHN; 'and if equal, equal ; and if less, less. Therefore,^ as the cir- cumference BC is to the circumference EF, so is the sedor BGC to the sedor EHF. Wherefore, in equal circles, kc, Q; E. D.- " 5. def. 5. i86 THE ELEMENTS Book VI. ""^^ PROP. B. THEOR. See N, 1 F an angle of a triangle be bisected by a straight line, which likewise cuts the base ; the rectangle contained by the sides of the triangle is equal to the rectangle contained by the segments of the base, together with the square of the straight line bisecting the angle. Let ABC be a triangle, and let the angle BAC be bise£led by the straight line AD ; the reftangle BA, AC is equal to the rei3:angle BD, DC, together with the square of AD. ' ^- *• Describe the circle* ACB about the triangle, and produce AD to the circumference in E, and join EC: Then because the angle BAD is equal to the angle CAE, and the angle ABD to the *2i.3. angle'' AEC, for they are in the samesegment; the triangles ABD, AEC, are equiangular to one ano- ther : Therefore as BA to AD, = 4.6. so is "^ EA to AC, and conse- quently the re£l:angle BA, AC is «• 16. 6. equaP to the redangle EA, AD, ' 3. 2. that is= to the redtangle ED, DA, together with the square of AD ; But the re£langle ED, DA * 35. 3. is equal to the redtangle*^ BD, DC. Therefore the re£langle BA, AC is equal to the redangle BD, DC, together with the square of AD. Wherefore, if an angle, &cc. Q^ E. D. PROP. C. THEOR. seeN. J^p f^QYYi any angle of a triangle a straight line be drawn perpendicular to the base ; the rectangle con- tained by the sides of the triangle is ecjual to the rectangle contained by the perpendicular and the diameter of the circle described about the triano-lc. 'O" Let ABC be a triangle, and AD the perpendicular from the angle A to the base BC j the re6langle B A, AC is equal to the rectangle contained by AD, and the diameter of the circle jde- scribed about the triangle. Describe OF EUCLID Describe* the circle ACB about the triangle, and draw its diameter AE, and join EC: Be- cause the right angle BDA is equal*" to the angle ECA in a se- micircle, and the angle ABD to the angle AEC in the same seg- ment"^ T the triangles ABD, AEC are equiano-ular : Therefore as «« BA to AD, so is EA to AC ; and consequently the retSbngle BA, AC is equal "^ to the red- 16.6. angle EA, AD. If therefore from an angle, &c. Q. E. D. PROP. D. THEOR. X HE rectangle contained bv the dia2:onals of asceN, quadrilateral inscribed in a circle, is equal to both the rectangles contained by its opposite sides. Let ABCD be any quadrilateral inscribed in a circle, and join AC, BD j the rectangle contained by AC, BD is equal to the two rectangles contained by AB, CD, and by AD, BC*. Make the angle ABE equal to the angle DBC ; add to each of these the common angle EBD, then the angle ABD is equal to the angle EBC : And the angle BDA is equal* to the angle BCE, because they are in the same segment : therefore the triangle ABD is equiangular to the triangle BCE: Wherefore'', as BC is to CE, so is BD to DA i and consequently the rectangle BC, AD is equal*^ to the rectangle BD, CE : Again, because the angle ABE is equal to the angle DBC, ,and the angle* BAE to the angle BDC, the triangle ABE is equi- angular to the triangle BCD : As therefore BA to AE, so is BD to DC; wherefore the rectangle BA, DC is equal to the rectangle BD, AE : But the rectangle BC, AD has been shewn equal to the rectangle BD, CE ; therefore the whole redtangl* AC, BD'' is equal to the rectangle AB, DC, together with the redangle AD, BC. Therefore the reaangle, &c. Q. E. D. ♦This is a Lemma of CI. Ptolomseus, in page 9, of iys u-c^aKn mret^n. 21. 3. 4.6. M6. 6. d 1, O. Book XL t x88 ] THE EL E M E N T S OF EUCLID. BOOK. XI. DEFINITIONS. I. J\ SOLID is that which hath length, breadth, and thickness, 11. That which bounds a solid is a superficies. III. A straight line is perpendicular, or at right angles to a plane, when it makes right angles with every straight line meeting it in that plane. IV. A plane is perpendicular to a plane, when the straight lines drawn in one of the planes perpendicularly to the common section of the two planes, are perpendicular to the other plane. V. The inclination of a straight line to a plane is the acute angle contained by that straight line, and another drawn from the point in which the first line meets the plane, to the point in which a perpendicular to the plane drawn from any point of the first line above the plane, meets the same plane. VI. The inclination of a plane to a plane is the acute angle con- tained by two straight lines drawn fr©m any the same point of their common section at right angles to it, one upon one plane, and the other upon the other. plane. VII. Two THE ELEMENTS OF EUCLID. 189 VII. Book XI. Two planes are said to have the same, or alike inclination to ^'^^' one another, which two other planes have, when the said aneles of inclination are equal to one another. VIII. Parallel planes are such which do not meet one another though produced. IX. A solid angle is that which is made by the meeting of more g^ ^ than two plane angles, which are not in the same plane, in one point. X. * The tenth definition is omitted for reasons given in the notes.' See N. XL Similar solid figures are such as have all their solid angles See N. equal, each to each, and which are contained by the same - number of similar planes. XIL A pyramid is a solid figure contained by planes that are con- stituted betwixt one plane and one point above it in which they meet. XIII. A prism is a solid figure contained by plane figures, of which two that are opposite are equal, similar, and parallel to one another : and the others parallelograms, XIV. A sphere is a solid figure described by the revolution of a se- micircle about its diameter, which remains unmoved. XV. The axis of a sphere is the fixed straight line about which the semicircle revolves. XVI. The centre of a sphere is the same with that of the semicircle. XVII. The diameter of a sphere is any straight line which passes through the centre, and is teraainated both ways by the su- perficies of the sphere. XVIII. A cone is a solid figure described by the revolution of a right angled triangle about one of the sides containing the right angle, which side remains fixed. If the fixed side be equal to the other side containing the right angle, the cone is called a right angled cone ; if it be less than the other side, an obtuse angled, and if greater, an acute angled cone, XIX. The THE ELEMENTS XIX. The axis of a cone is the fixed straight line about which the triangle revolves. XX. The base of a cone is the circle described by that side contain- ing the fight angle, which revolves. XXI. A cylinder is a solid figure described by the revolution of a right angled parallelogram about one of its sides wrhich remains fixed. XXII. The axis of a cylinder is the fixed atraight line about which the parallelogram revolves. XXIII. The bases of a cylinder are the circles described by the two revolving opposite sides of the parallelogram. XXIV. Similar cones and cylinders are those which have their axes and the diameters of their bases proportionals. XXV. A cube is a solid figure contained by six equal squares, XXVI. A tetrahedron is a solid figure contained by four equal and equilateral triangles. XXVII. An octahedron is a solid figure contained by eight equal and equilateral triangles. XXVIII. A dodecahedron is a solid figure contained by twelve equal pentagons which are equilateral and equiangular. XXIX. An icosahedron is a solid figure contained by twenty equal and equilateral triangles. DEF. A. . A parallelopiped is a solid figure contained by six quadrilate- ral figures, whereof every opposite two are parallel. OF EUCLID. 191 Book XI. PROP. I. THEOR. UnE part of a straight line cannot be in a plane, seex. and another part above it. If it be possible, let AB, part of the straight line ABC, be in the plane, and the part BC above it : And since the straight line AB is in the plane, it can be produced in that plane: Let it be propucedtoD: And let any plane pass through the straight line AD, and be turned about it un- til it pass through the point C ; and because the points B, Care in this plane, the straight line BC is in it^: Therefore there are two straight lines ABC, * ^dcf-l. ABD in the same plane that have a common segment AB, which is impossible''. Therefore, one part, &c. Q. E. D. i»Cor. 11.1 PROP. n. THEOR. 1 WO straight" lines ■which cut one another are in one plane, and three straight lines which meet one another arc in one plane. Let two straight lines AB, CD, cut one another in E ; AB» CD are one plane : and three straight lines EC, CB, BE^ which meet one another are in one plane. Let any plane pass through the straight line EB, and let the plane be turned about EB, produced, if necessary, until it pass through the point C : Then because the points E, C are in this plane, tthe straight line EC is in it*: For the same ^y\ ' '^^*'"'' '' reason, the straight line BC is in the same; and, by the hypothesis, EB is in it : Therefore the three straight lines EC, CB, BE are in one plane : But in the plane in which EC, EB are, in the same are" CD, AB: Therefore AB, CD, are j. ,;, in one plane. Wherefore two straight lines, &c. Q^ E. B. THE ELEMENTS PROP. III. THEOR. seeN. JLF two plaiics cut onc another, their common sec- tion is a straight Hne. Let two planes AB, 'BC, cut one another, and let the line DB betiieir common section: DB is a straight line : If it be not, from the point p to B, draw, in the plane AB, the straight line DlEB, and in the plane BC, the straight line DFB : Then two straight lines DEB, DFB have the same extremities, and therefore include a space *jOAx. 1. betwixt them; which is umpossibie^ : Therefore BD the common section of the planes AB, BC, cannot but be a straight line. Wherefore, if two planes, &c. Q, E. D. .B E DN A^ SeeN. « 15. 1. ^ 4. J. « 26. I. PROP. IV. THEOR. IF a straight line stand at right angles to each of two straight lines in the point of their intersection, it shall also he at right angles to the plane which passes through them, that is, to the plane in which they are. ■ Let the straight line EF stand at right angles to each of the straight lines AB, CD in E, the point of their intersection : EF is also at right angles to the plane passingthrough AB, CD. Take the straight lines AE, EB, CE, ED,'all equal to one an- other; and through E draw, in the plane in which are AB,CD, any straight line GEH ; and join AD, CB ; then, from any point F in EF, draw FA, FG, F'D, FC, FH, FB : And because the two straight lines AE^ ED are equal to the two^BE, EC, and that they contain equal angles^ AED, BEC, the base AD, is equal'' to the base BC, and the angle DAE to the angle EBC : And the angle AEG is equal to the angle BEH"; there- fore the triangles AEG, BEH have two angles of one equal to two angles of the other, each to eaeh. and the sides AE, EB, adjacent to the equal angles, equal to onc another : where- fore they shall have their other sides equal"^ : GE is therefore equal OF EUCLID. ^93 b-i. 1. s. 1. vaual to EH, and AG to BH: and because AE is equal to ^°°^J^' EB, and FE common and at rightangles'to them, the base AF ^■'^'''^■^. is equal*" to the base FB ; for the same reast a, CF is equal to FD: And because AD is equal to BC, and AF to FB, the two sides FA, AD are equal to the two FB, BC, each to each j and the base p DF was proved equal to the base FC ; therefore the angle FAD is equal " to the angle FBC : Again, it was proved that GA is equal toBH, and also AF A.^ to FB ; FA, then, and AG, are equal to FB and BH, and the angle FAG has been proved equal to the angle FBH ; therefore the base GF is equal ''to the base FH : Again, because it was proved, that GE is equal to EH, and EF is common ; GE, EF are equal to HE, EF ; and the base GF is equal to the base FH ; therefore the angle GEF is equal' to the angle KEF ; and consequently each of these angles is a right^ angle. Therefore FE makes right angles with GH, * lOdef. l. that is, with any straight line drawn through E in the plane passing through AB, CD. In like manner, it may be proved, that FE makes right angles with every straight line which meets it in that plane. But a straight line is at right angles to a plane when it makes right angles with every straight Tine , which meets it in that plane*": Therefore EF is at right an- gles to the plane in v/hich are AB, CD. straight line, 5cc. Q^ E D. Wherefore, if a ' 3 d«i; 11. PROP. V. THEOR. JlF tliree straight lines meet all in one point, and aseex. straight line stands at right ang;les'to each of them in that point ; tliese three straight lines are in one and the same plane.' Let the straight line' AB stand at right angles to each of the straight lines BC, BD, BE, in B the point where they meet; BC, BD, BE, are in one and the same plane. If not, let, if it' be possible, BD and BE be in one plane, and BC be above it ; and let a plane pass through AB, BC, he common section of which v/ith the plane, in which BD O and 19+ THE ELEMENTS M. II. Book x^, a^j BE are, shall be a straight*^ line ; let this be BF: There- ^"sAL ^°^^ ^^^ three straight lines AB, BC, BF, are all in one plane, viz. that which passes through AB, BC; and because AB stands at right angles to each of the straight lines BD, BE, it is also at right angles ^ to the plane passing through them i ^3 def. 11. ^pj therefore makes right angles'^ with every straight line meeting it -A. ill that plane ; but BF which is in that plane meets it : Therefore the angle ABF is a right angle ; but the angle ABC, by the hypothesis, is also a right angle ; therefore the angle ABF is equal to the angle ABC, and they are both in the same plane, which is impossible ; Therefore the straight line BC is not above the plane in which are BD and BE : Wherefore the three straight lines BC, BD, BE are in one and the same plane. Therefore, if three straight lines, &c. Q. E. D. PROP. VI. THEOR. 1 F two straight lines be at right angles to the same plane, they shall be parallel to one another. Let the straight lines AB, CD be at right angles to the, same plane ; AB is parallel to CD. Let them meet the plane in the points B, D, and draw thij straight line BD, to which draw DE at right angles, in thel same plane; and make DE equal to AB, and join BE, AE, AD. Then, because AB is perpendicular to the plane, it *. 3 def. 11. shall make right' arjgles with every straight line which meets it, and is in that plane : But BD, BE, which are in that plane, do each of them meet AB. Therefore each of the angles ABD, ABE is a right angle : For the same rea- son, each of the angles CDB, CDE is a right angle : And because AB is equal to DE, and BD common, the two sides AB, BD are equal to the two ED, DB ; and tljey contain right angles ; therefore the base AD i» equal'' to thq base BE: Again, because AB is equal M. }. t« OF EUCLID. ' «95 to DE,and BE to AD ; AB, BE are equal to ED, DA; and, ^;^*^- in the triangles ABE, EDA, the base AE is common : there- ^^"'■^^'^^ fore the angle ABE is equal <= to the angle EDA: But'8-i- ABE is a right angle : therefore EDA is also a right angle, and ED perpendicular to DA : But it is also perpendicular to each of the two ED, DC : Wherefore ED is at right angles to each of the three straight lines BD, DA, DC in the point in which they meet : Therefore these three straight lines are all in the same plane"^ : But AB is in the plane in which are BD, * ^- i^- DA, because any three straight lines which meet one another are in one plane*^ : Therefore AB, BD, DC are in one plane ; * 2. ij| And each of the angle's AED, BDC is a right angle; therefore AB is parallel^ to CD. Wherefore, if two straight lines, &c. '^^S. 1. Q,E. D. PROP. VII. THEOR. AF two straight lines be parallel, the straight line See2*» ' drawn from any point in the one to any point in the other, is in the same plane with the parallels. Let AB, CD be parallel straight lines, and take any point E in, the one, and the point F in the other: The straight line which joins E and F is in the same plane with the parallels. If not, let it be, if possible, above the plane, as EGF ; and in the plane ABCD in which the parallels are, draA^ the straight A "F "R line EHF from E to F ; and since - . . *M EGF also is a straight ime, the two straight lines EHF, EGF include a space between them, ■which is impossibles There- \] »iOA.t. i. fore the straight line joining the C t D points E, F is not above the plane in which the parallels AB, CD are, and is therefore in that plane. Wherefore, if two straight lines, &c. Q. E. D. PROP.VIIL THEOR. XF two straight lines be parallel, and one of them set is at right angles to a plane ; the other also shall be at right angles to the same plane 02 I^ JT'T. II. THEELEMENTS Let AB, CD be two parallel straight lines, and let one of them AB be at right angles to a ^lane ; the other CD is at right angles to the same plane. Let AB, CD meet the plane in the- points B, D, and join BD ; Therefore » ABj CD, BD are in one plane. In the plane to which AB is at right angles, draw DE at right angles to BD, and make DE equal to AB, and join BE, AE, AD. And because AB is perpendicular to the plane, it is perpen- dicular to every straight line which meets it, and is in that »3. del. 11. plane^ : Therefore each of the angles ABD, ABE, is a right angle : And because the straight line BD meets the parallel straight lines AB, CD, the angles ABD, CDB are''together '.'r>. 1. . aqual*" to two right angles: And ABD is a right angle; therefore also CDB is a right angle, and CD perpendicular to BD : And because AB is equal to DE, and BD common, the two AB, BD are equal to the two ED, DB, and the angle ABD is equal to j^^ the angle EDB, because each of them is a right angle ; therefore the" base AD ' ' '• is equal <= to the base BE: Again, because AB is equal to DE, and BE to AD ; the two AB, BE, are equal to the two ED, DA j and the base AE is com- mon to the triangles ABE, EDA; wherefore the angle ABE is equal ^ to the angle EDA : And ABE is a right angle; and therefore EDA is a right angle, and ED perpendicular to DA: But it is also perpendicular to BD ; therefore ED is perpen- dicular'^ to the plane which passes through BD, DA, and shall*" make right angles with every straight line meeting it in that plane : But DC is in the plane passing through BD, DA, be- cause all three are in the plane in which are the parallels AB, CD : " Wherefore ED is at right angles to DC ; and therefore CD is at riglit angles to DE : But CD is also at right angles to DB ; CD then is at^ right "angles to. the two straight lines DE, DB in the point of theii* intersection D : and therefore is at right angles'" to the plane passing through DE, DB, which is the same plane to which A B is at right angles. Therefore, if two straight jjftes, &c. ^C^E..D. "8. 1, -4. n. 'Jdef. 11 OF EUCLID. PROP. IX. THEOR. X WO straight lines which are each of them paral- lel to the same straight line, and not in the same plane with it, are parallel to one another. Let AB, CD, be each of them parallel to EF, and not in the same plane with it; AB shall be parallel to CD, In EF cake any point G, from which draw, in the plane passing through EF, AB, the straight line GH at right angles to EF ; and in the plane passing through EF, CD, draw GK atrightanglesto the same EF. And because E^F is perpendicular both, to Gri and GK, EF is perpendi- cular* to the plane HGK passing through them : and EF is parallel toAB; therefore AB is at right angles *= to the plane HGK. For the same reason, CD is likewise at r? right angles to the plane HGK. Therefore AB, CD are each of them at right angles to the plane HGK. But if two straight lines are at right angles to the same plane, they shall be parallel*^ to one another. There- fore AB is parallel to CD. Wherefore, two straight lines, &c. Q, E. D. - A H rh \ J5 F D »4. 11. 8.11. 6. n. PROP. X. THEOR. J F two straight lines meeting one another he pa- rallel to two others that meet one another, and are not in the same plane with the first two ; the first two and the other two shall -contain equal angles. •Let the two straight lines AB, BC, which meet one ano- ther, be paralleLto the two straight lines DE, EF that meet •ne another, and are not in the same plane with AB, BC. The angle ABC is equal to the angle DEF. Take BA, BC, ED, EF all equal to one another j and join O 3 AD, BookXI. '35. 1. og. 11. = l.Ax. 1. ■is. 1. THE ELEMENTS AD, CF, BE, AC, DF : Because BA is equal and parallel to ED, therefore AD is* both equal and parallel to BE. For the same reason, CF is equal and parallel to BE. There- fore AD and CF are each of them equal and parallel to BE. But straight lines that are parallel to the same straight line, and not in the same plane with it, are parallel'' to one another. Therefore AD is parallel to CF ; and it is equal*: to it, and AC, DF join them towards the same parts ; and therefore ' AC is equal and parallel to DF. And be- cause AB, BC are equal to DE, EF, and the base AC to the base DF ; the angle ABC is equaH to the angle DEF.. Therefore, if two straight lines, &c. Q. E. D. *12. 1. bli. I. '31. 1. '4. 11. «8. 11. *3def. 11. PROP^^I. PROB. 1 O draw a straight line perpendicular to a plane, from a given point above it. Let A be the given point above the plane BH ; it is required to draw from the point A a straight line perpendicular to the plane BH. In the plane draw any straight line BC, and from the point A draw ^ AD perpendicular to BC. If then AD be also per- pendicular to the plane BH, the thing required is already done ; but if it be not, from the point Ddraw'*, in the plane BH, the straipjht line DE, at right an- gles to BC ; and from the point A draw AF perpendicular to DE ; and through F draw"^ GH parallel to BC : And because BC is at right angles to ED and DA, BC is at right angles'* to the plane passing through ED, DA. And GH is parallel to RC ; but, if two straight lines be parallel, one of which is at right angles to a plane, the other shall be at right <^ angles to the same plane ; wherefore GH is at right angles tothe plane through ED,DA, and is perpendicular ^ to every straight line meeting it in that plane. But AF, which is in the plane through ED, AD, meets it: OF EUCLID. ,99 it : Therefore GH Is perpendicular to AF ; and consequently ^ook XI. AF is perpendicular to GH ; and AF is perpendicular to DE: ^*^V"**^ Therefore AF is perpendicular to each of the straight lines GH, DE. But if a straight line stands at right angles to each of two straight lines in the point of their interseclionj i: shall also be at right angles to the plane passing through them. But the plane passing through ED, GH is the plane BH ; therefore AF is perpendicular to the plane BH ; therefore, from the given point A, above the plane BH, the straight line AF is drawn perpendicular to that plane : Which was to be done. PROP. XII. PROB. X O erect a straight line at right angles to a given plane, from a point given in the plane. Let A be the point given in the plane ; it is required re£t a straight line from the point A at right ^^ ^ angles to the plane. -^ From any point B above the plane draw* BC perpendicular to it ; and from A draw^ AD parallel to BC. Because, therefore, AD, CB are two parallel straight lines, and one of them B *ii. 11. »31. U BC is at right angles to the given plane, / j\ f! / the other AD is also at right angles to it^ Therefore a straight line has been erected at right angks c g. ^ to a given plane from a point given in it. Which was to be done. PROP. Xm. THEOR. -T ROM the same point in a given plane, there cannot be two straight lines at right angles to the plane, Bpon the same side of it ; and there can be but one per- pendicular to a plane from a point above the plane. For, if it be possible, let the two straight lines AB, AC, be at right angles to a given plane from the same point A in» the plane, and upon the same side of it; and let a plane pass through B Aj AC i the common section of this with the given plane is a O 4 straight 200 Book XI. ^S. U. 6. 11. THE ELEMENTS straight* line passing through A: Let DAE be their common se£lion: Therefore the straight lines AB, AC, DAE are in one plane : And because CA is at right angles to the given plane, it shall make right angles with every straight line meetingit inthatplane. But DAE, which is in that plane, meets CA ; therefore CAE is a right angle. For the same reason BAE is a right angle. Wherefore the an- gle CAE is equal to the angle BAE; and they are in one plane, ■which is impossible. Also, from a point above a plane, there can be but one perpendicular to that plane ; for, if there could be two, they would be parallel'' to one another, which is ab- surd. Therefore, from the same point, &c. Q. E. D. PROP. XIV. THEOR. ■■3M. 1 «• 17. 1. 'S(ief. U. 1 LANES to which tli^ same straight line is per- pendicular, are parallel to one another. Let the straight line AB be perpendicular to each of the planes CD, EP' s these planes are parallel to one another. If not, they shall meet one another when produced ; let them meet ; their common sed"ion shall be a straight line GH, in which take any point K, and join AK, BK : Then, because AB is perpendicular to the plane EF, it is perpendicular" to the straight line BK which is in that plane. Therefore ABK is a right angle. For the same reason BAK is a right angle ; wherefore the two angfes ABK, BAK of the triangle ABK are equal to two right, angles, which is impossible'' ; Therefore the planes CD, EP\ though produced, do not meet one another j that is, they are parallel. There- fore planes, &c. Q. K. D. OF EUCLID. PROP. XV. THEOR. 51. I. «3dctll. jlF two straiglil lines meeting one another, be pa- sceN. rallel to two straight lines which meet one another, but are not in the same plane with the first two ; the plane M-hich passes through these is parallel to t he plane passing through the others. Let AB, BC, two straight lines meeting one another, be parallel to DE, EF that meet one another, but are not in the same plane with i\B, BC : The planes through AB, BC, and D£, EF shall not meet, though produced. From the point B draw BG perpendicular* to the plane * n. n. which passes through DE, EF, and let it meet that plane in Gj and through G draw GH parallel'' to ED, and GK parallel to EF : And because BG is perpendicular to the plane through DE, EF, it shall make right angles with every straight line meeting it in that planed But the straight lines GH, GK ill that plane meet it: Therefore each of the angles BGH, BGK is a right angle : And because BA is parallel'' to GH (for each of them is parallel to DE, and they are rtot both in the same plane with it) the angles GBA, BGH are together equa^ to two right angles : And BGH is ' 29, i. a right angle ; therefore also GBA is a right angle, and GB perpendicular to BA : For the same reason, GB is perpendi- cular to BC : Since therefore the straight line GB stands at right angles to the two straight lines BA, BC, that cut one another in B ; GB is perpendicular* to the plane through BA, f^ ij_ BC : And it is perpendicular to the plane through DE, EF j therefore BG is perpendicular to each of theplanes through AB, 8C, and DE, EF : But planes to which the same straight line perpendicular, are parallel? to one another : Therefore the g f^, n plane through AB, BC is parallel to the plane through DE, EF. Wherefore if two straight lines, &c. Q. E. D. See N. THE ELEMENTS PROP. XVI. THEOR. JlT two parallel planes be cut by another plane, their common sections with it are parallels. Let the parallel planes, AB, CD be cut by the plane EFHG, and let their common sections with it be EF, GH : EF is parallel to GH. For, if it is not, EF, GH shall meet, if produced, either on thesideof FHjOr EG : First, let them be produced on the side of FH, and meet in the point K : Therefore, since EFK is in the plane AB, every point in EFK is in that plane ; and K ^3 a point in EFK ; therefore K is in the plane AB : For the same reason K is also in the plane CD : Wherefore the planes, AB, CD, produced meet one another ; but they do not meet, since they are parallel by the hypothesis : Therefore, the straight lines EF, GH do not meet when produced on the side of FH : In the same manner it may be proved, that EF, GH do not meet when produced on the side of EG : But straight lines which are in the same plane, and do not meet, though produced either way, are parallel : There- fore EF is parallel to GH. Wherefore, if two parallel planes, &c. Q. E. D. D PROP. XVn. TIIEOR. If two straight lines be cut by parallel planes, they shall be cut in the same ratio. Let the straight lines AB, CD be cut by the parallel planes GH, KL, MN, in the points A, E, B j C, F, D : As AE Hi to EB, so is CF to FD. Join AC, BD, AD, and let AD meet the plane KL in th< point X: and join EX, XF: Because the two parallel plane^ KL, MN are cut by the plane EBDX, the common se^ion? EX, OF EUCLID. 203 EX, BD are prirallels For the same reason, because the two ^*«^ ^ parallel planes GH, KL are cut 1 }tj_ j 1^ by the plane AXFC, the com- ^__ \\ mon sections AC, X Fare parai- / a lel : And because EX is paral- qL — L- lel to BD, a side of the triangle ABD, as AE to EB, so is^ AX to XD. Again, because XF is parallel to AC, a side of the triangle ADC, as AX to XD, so is CF to FD: And it was ^^ proved that AX is to XD, as AE to EB : Therefore% as I ^^_\]^ xr MI.5. AE to EB, so is CF to FD. V/herefore, iftwostraightlines, \\^ &c. Q. E. D. PROP. XVIII. THEOR. F a straight line be at right aagles to a plane, every plane which passes throu^ it shall Jjje at right an- gles to that jjlane. Let the straight line AB be at right angles to a plane CK ; every plane which passes chrovgh AB shall be at right angles to the plane CK. Let any plane DE pass through AB, and let CE be the com- I mon section of the planes DE, CK; take any point F in CE, 'rn which draw FG in the -ae DE at right angles to CE : And because AB is per- pendicular to the plane CK, therefore it is also perpendi- cular to every straight line in that plane meeting it^: And \ \ \ »3def, 11, consequently it is perpendicu- lar to CE : Wherefore ABF is a right angle ; But GFB is ■ D G A Tf K \ \ \ \ F B E likewise a right angle ; therefore AB is parallel'' to FG. And b og. j, AB is at right angles to the plane CK ; therefore FG is also at right angles to tiie same planed But one plane is at right an- c 3. n. js to another plane when the straight lines drawn in one of the -nes, at right angles to their common section, are also at right angles ,204 THE ELEMENTS Book XI. anglcs to the other plane' ; and any straight line FG in the d i def. 11. plane DE, which is at right angles to CE the common sedtion of the planes, has been proved to be perpendicular to the other plane CK ; therefore the plane DE is at right angles to the plane CK. In like manner, it may be proved that all the planes which pass through AB are at right angles to the plane CK. Therefore, if a straight line. Sec. Q^ E. D. PROP. XIX. THEOR. XF two planes cutting one another be each of them perpendicular to a thud plane ; their common sec- tion shall be perpendicular to tlie same plane. Let the two planes AB, BC be each of them perpendicular to a third plane, and let BD be the commen section of the first two ; BD is perpendicular to the third plane. If it be not, from the point D draw, in the plane AB, the straight line DE at right angles to AD the common se6lion of the plane AB with the third plane ; and in the plane BC draw DF at right angles to CD the common sedlion of the plane BC with the third'plane. And because the plane AB is perpendicular to the third plane, and DE is drawn in the plar.e AB at right angles to AD their com- mon section, DE is perpendicular to + def. 11. the third plane*. In the same manner, • it may be proved that DF is perpen- the third plane. Where- the point D two straight at right angles to the third same side of it, which Therefore, from the fci3. n. dicular to fore, from lines stand plane, upon the is impossible'' point D there cannot be any straight line at right angles to the third plane, except BD the common section of the planes AB, BC. Bl therefore is perpendicular to the third plane. Wherefore^ ii two planes, &c. Q. E. D. OF EUCLID. 205 Bock XI. PROP. XX. THEOR. "^'^^ I Fa solid angle be contained by three plane an-'s« n. gles, any two of them are greater than the third. Let the solid angle at A be contained by the three plane angles, BAG, CAD, DAB. Any two of them are greater than the third. it' the angles BAC, CAD, DAB be all equal, it is evident that any two of them are greater than the third. But if they are not, let BAC be that angle which is not less than either of the other two, and is greater than one of them DAB ; and at the point A iij the straight line iVB, make, in the plane which passes through BA, AC, the angle BAE equal^ to the angle »23. 1. DAB ; and make AE equal to AD, and through E draw EEC cutting AB, AC in the points B, C, and join DB, DC. And be- cause DA is equal to AE, and AB is common, the two DA, AB are equal to the two EA, AB, and the angle 'Dab is equal to the angle EAB : Therefore the base DB is equal* to / --^" "^^^^^^^^^-^^ — \ b-i. 1. the base BE. And because BD, DC are greater "= than CB, and one of "*' "^^ ^ c 00. 1. them BD has been proved equal to BE a part of CB, therefore the other DC is greater than the remaining part EC. And be- cause DA is equal to AE, and AC common, but the base DC greater, than the base EC ; therefore the angle DAC is greater** <i 25. 1. than the angle E AC ; and, by the constru£tion, the angle DAB jis equal to the angle BAE ;* wherefore the angles DAB, DAC I are together greater than BAE, EAC, that is, than the angle I BAC. But^BAC is noMess than either of the angles DAB, iDAC ; therefore BAC, with either of them, is greater than 'the other. Wherefore if a solid angle, &c. Q. E. D. PROP. XXL THEOR. ^-tliVERY solid anajie is contained bv plane auo-les, [which together are less than four right angles. First, Let the solid angle at A be contained by three plane jangles BAC, CAD, DAB. These three together are less ithan four right angles. ? Take 206 Book XI. ^'QO. 11. > 32. 1. THE ELEMENTS Take in each of the straight lines AB, AC, AD any points B, C, D, and join BC, CD, DB : Then, because the solid angle at B is contained by the three plane angles CBA, ABD, DBG, any two of them are greater^ than the third ; therefore the angles CBA, ABD are greater than the angle DBC: For the same reason, the angles BCA, ACD are greater than the angle DCB ; and the angles CDA, ADB greater than BDC : Where- fore the six angles CBA, ABD, JiCA, ACD, CDA, ADB are greater than the three angles DBC, BCD, CDB : But the three angles DBC, BCD, CDB are equal to two right angles'' : Therefore the six angles CBA, ABD, BCA, ACD, CDA, ADB are greater than two right angles : And because the three angles of each of the triangles ABC, ACD, ADB are j>~ equal to two right angles, therefore the nine angles of these three triangles, viz. the angles CBA, BAC,ACB,ACD, CDA, DAC, ADB, DBA, BAD are equal to six right anf^les: Of these the six angles CBA, ACB, ACD, CDA, ADB, DBA are greater than two right angles : Therefore the remaining three angles BAC, DAC, BAD, which contain the solid angle at A, are less than four right angles. • Next, Let the solid angle at A be contained by any number of plane angles BAC, CAD, DAK, EAF, FAB : these together are less than four right angles. Let the planes in which the angles are, be cut by a plane, and let the common sections of it with those planes be BC, CD, DE, EF, FB : And because the solid angle at B is contain- ed by three plane angles CBA, ABF, FBC, of which any two are greater^^ than the third, the angles CBA, ABF are greater than the angle FBC : for the same reason, the two plane angles at each of the points C, D, E, F, viz. the angles which are at the bases of the triangles having the common vertex A, are greater than the third angle at the same point, which is one of the an- gles of the polygon BCDEF : There- fore all the angles at the bases of the triangles are together. greater OF EUCLID. 207 greater than all the angles of the polygon : And becausa all the Book xi. angles of the triangles are together equal to twice as many ""^^ right angles as there are triangles^ ; that is, as there are sides » Z2. 1. in the polygon BCDEF ; and that all the angles of the poly- gon, together with four right angles, are likewise equal to twice as many right angles as there are sides in the polygon*^ ; * '• ^°''* therefore all the angles of the triangles are equal to all the angles of the polygon together with four right angles. But all the angles at the bases of the triangles are greater than all the angles of the polygon, as has been proved. Wherefore the remaining angles of the triangles, viz. those at the vertex, which contain the solid angle at A, are less than four right an- gles. Therefore every soUd angle, &c. Q. E. D. PROP. XXII. THEOR. JlF every two of three plane angles be greater than s^^"- the third, and if the straight lines which contain tliem be all equal ; a triangle may be made of the straight lines that join the extremities of those equal straight lines. Let ABC, DEF, GHK be three plane angles, whereof every two are greater than the third, and are contained bv the equal straight lines AB, BC, DE, EF, GH, HK ; if their extremities be joined by the straight lines AC, DF, GK, a tri- angle may be made of three straight lines equal to AC, DF, GK ; that is, every two of them are together greater than the third. If the angles at B, E, H, are equal, AC, DF, GK are also (equal* and any two of them greater than the third : But if**- ^• the angles are not all equal, let the angle ABC be not less than jeither of the two at E, H ; therefore the straight line AC is not less than either of the other two DF, GK" 3 and it is " 4. Cor. plain that AC, together with either of the other two, must be ^*' '• greater than the third : Also DF with GK are greater than AC : For, at the point B, in the straight line AB, make'' the <^q3. 1. ande o8 THE ELEMENTS Book XI. angle ABL equal to the angle GHK, and make BL equal to ^-^'^^ one of the straight lines AB, BC, DE, EF, GH, HK, and join AL, LC ; Then, because AB, BL are equal to GH, HK, and the angle ABL to the angle GHK, the base AL is equal to the base GK : And because the angles at E, Hare greater than, the angle ABC, of which the angle at H is equal to ABL, there- fore the remaining angle at E, is greater thau the angle LBC : ScfN. F G And because the two sides LB, BC are equal to, the two DE, EF, and that the angle DEE is greater than the angle LBC, the * 24. 1. base DF is greater'' than the base LC: And it has been proved that GK is equal to AL ; therefore DF and GK are greater ' 20. 1. than AL and LG: But AL and LC are greater^ than AC j much more then areDF andGKgreater than AC. Whereroreevciy two of these straight lines AC, DF, GK are greater than the •^32.1. third; and, therefore, a triangle may be made*^, the sides of which shall he equal to AC, DF, GK. Q. E. D. PROP. XXIIL PROB. X O make a solid angle which shall be contained by three given plane angles, any two of them being ii-reater than the third, and all three too'Cther less than four riglit angles. Let the three given plane angles be ABC, DEF, GHK,' any two of which are greater than the third, and all of thenif together less then four right angles. It is required to make a 'V solid angle contained by three plane angles equal to ABC, * DEF, GHK, each to each. . - From OF EUCLID. 209 From the straight lines containing the angles, cut ofF AB ; Book XI. BC, DE, EF, Gil. HK, all equal to one another j and join ""^'"^ AC, DFj GK : Then a triangle may be made' of three straight » 22. 11. 22. 1. 5. 4. lines equal to AC, DF, GK. Let this be the triangle LMN", so that AC be equal to LM, DF to MN, and GK - to LN i and about the triangle LMN describe*^ a circle, and c find its centre X, which will either be within the triangle, or in one of its sides, or without it. First, Let the centre X be within the triangle, and join LX, MX, NX : AB is greater than LX : If not, AB must either be equal to, or less than LX ; first, let be equal : Then because AB is equal to LX, and that AB is also equal to BC, and LX to XM, AB and BC are equal to LX and XM, each to each ; and the base AC is by construdlion, equal to the base LM 3 wherefore the angle ABC is equal to the angle I LXM**. For the same reason, the angle DEF is equal to the <i 8. 1. langle MXN, and the angle GHK to the angle NXI, : Therefore the three angles ABC, DEF, GHK are (equal to the three angles LXM, jMXN, NXL: But the three angles [LXM, MXN, NXL are equal to Ifour right angles'^; therefore also Oie three andes ABC, DEF, GHK ir^equal to four right angles : But, >y the hypothesis, they are less vf^ lan four right angles, which is absurd j therefore AB is not equal to LX: But neither can AB be ess than LX : For, if possible, let it be less, and upon the raight line LM, on the side of it on which is the centre X, -scribe the triangle LOM, the sides LO, OM of which are iual to AB, BC ; aud because the base LM is equal to the P base 2IO THE ELEMENTS ^°^^^- based AC, the angle LOM is equal to the angle ABC^ : And 4 sPfT*'^ ■A-Bj *^^^ ^^' LO, by the hypothesis, is less than LX; where- fore LO, OM tail within the triangle LXAl j for, if they fell upon its sides, or without it, they would be equal to or greater than ^21.1. LX, XM^: Therefore the angle LOM, that is, the angle ABC, is greater than the angle LXM^: In the same manner it may be proved that the angle DEF is greater than the angle MXN, and the angle GHK greater than the angle NXL. Therefore the three angles ABC, DEF, GHK are greater than the three angles LXM, MXN, NXL ; that is, than four right angles : But the same angles ABC, DEF, GHK are less than four right angles i which is absurd : Therefore AB is nofless than LX, and it has been proved that it is not equal to LX ; wherefore AB is greater than LX. Next, Let the centre X of the circle fall in one of the sides of the triangle, viz. in MN, and » join XL : In this case also AB is greater than LX. If not, AB is either equal to XL, or less than it : First, ]?t it be equal to XL : There- fore AB and BC, that is, DE, and EF, are equal to MX and XL, that iSjtoMN: Butjby the construction, MN is equal toDF; therefore DE, M'f EF are equal to DF, which is im- t20. 1. possiblef: Wherefore A B is not , equal to LX; nor is it less; for then, much more, an absurdity would follow : Therefore AB is greater than LX. But, let the centre X of the circle fail without the triangle LMN, and join LX, MX, NX. In this case likewise AB is i greater than LX : If not, it is either equal to, or less than LX: j First, let it be equal ; it may be proved in the same manner, i, as in the first case, that the angle ABC is equal to the angle MXL,and GHK to LXN; therefore the whole angle MX>f is equal to the two angles, ABC, GHK : But ABC and GHK are together greater than the angle DEF ; therefore also the angle MXN is greater than DEF, And because DE, EF OF EUCLID. 211 EF are equal to MX, XN, and the base DF to the base ^ook. xi- MN, the angle MXN is equal"^ to the ang]e DEF : And it has ^^f*^ been proved, that it is greater than DEF, which is absurd. Therefore A B is not equal to LX. Nor yet is it hss ; for then, as has been proved in the first case, the angle ABC is greater than the angle MXL, and the angle GHK greater than the angle LXN . At the point B, in the straight lineCB,make the angle CBP equal to the angle GHK, and make BP equal to / / G~ HK, and join CP, AP. And because CB is equal to GH ; CB, BP are equal to GH, HK, each to each, and they contain equal angles; wherefore the base CP is equal to the base GK, that is, to LN. And in the isosceles triangles ABC, MXL, because the angle ABC is greater than the angle MXL, there- fore the angle MLX at the base is greater 5 than the angle t sg AC Bat the base. For the same reason,because the angle GHK, or CBP, is greater than the angle LXN, the angle XLN is greater than the angle BP. Therefore the whole angle MLX is greater than the whole angle ACP. And because ML, LN are equal to AC, CP, each to each, but the angle MLN is greater than the angle ACP, the base MN is greater ^ than the base AP. And MN is equal to DF ; therefore also DF is greater than AP. Again, because DE, EF are equal to AB, BP, but the baseDF greater than the base AP,the angle DEF is greater'' than the angle ABP. And ABP is equal to the two angles ABC, CB", that IS, to the two angles ABC, GHK ; therefore the angle IJEF is greater than the two angles ABC, GHK ; "but it is also less than these, which is impossible. Therefore AB h not less than P2 LX; »24. 1. 2J. I. 212 THE ELEMENTS Book XI. LX, and it has been proved that it is not equal to it ; there- ^"^^^''^^ fore AB is greater than LX. « 12. 11. From the point X ere6l* XR at right angles to the plane of the circle LMN. And because 't has been proved in all the cases, that AB is greater than LX, find a square equal to the excess of the square of AB above the square of LX, and make HX "* equal to its side, and join RL,RM, RN. Because RX is perpendicular to the plane of the circle LMN, it " Sdef. 11, is'' perpendicular to each of the straight lines LX, MX, NX And because LX is equal to MX, and XR common, and at right angles to each of tbem, the base RL is equal to the base RM. For the same reason, RN is equal to each of the twoRLjRM. 1 hereforethethree straight lines RL,RM,RN,areaIl equal. And becausp the square of XR is equal to the excess of the square of AB above the square of LX ; therefore the square of AB is equal to the squares of >- 47, 1. LX, XR. But the square of RL is equal' to the same squares, because LXR is a right angle. Therefore the square of AB is equal to the square of RL, and the straight line AB to RL. But each of the straight lines BC, DE, EF, GH, HK is equal to AB, and each of the twoRM,RN is equal to RL. Where- fore AB, BC, DE, EF, GH, HK, are each of them equal to each of the straight lines RL, RM, RN. And because RTv, RM, are equal to AB, BC, and the base LM to the base AC ; « 8. 1, the angle LRM is equal*^ to the angle ABC. For the same reason, the angle MRN is equal to the angle DEF, and NRL to GHK. Therefore there is made a solid angle at R, which is contained by three plane angles LRM, MRN, NRL, which are equal to the three given plane angles ABC, DEF, GHK, each to each. Which was to de done. OF EUCLID. PROP. A. THEOR. If each of two solid angles be contained by three s^ex. plane angles equal to one another, each to each ; the planes in which the equal angles are to have the same incUnation to one another. Let there be two solid angles at the points A, B ; and let the ano-le at A be contained by the three plane angles CAD, CAE,^AD ; and the angle at B by the three plane angles FBG FBH HBG ; of which the angle CAD is equal to the angle FBG,' and CAE to FBH, and EAD to HBG : The planes in which the equal angles are, have the same inclination to one another. In the straight line AC take any point K, and in the plane CAD from K draw the straight line KD at right angles to AC, and in the plane CAE the straight line KL A. at right angles to the same AC : Therefore the angle DKL is the inclination * of the plane CAD to the plane CAE: In BF take BiVl equal to AK, and from the point M draw, in the planes FBG, FBH, the straight lines MG, MN at right angles to BF ; therefore the angle GMN is the incli- nation ^ of the plane FBG to the plane FBH : Join LD, NG; and because in the triangles KAD, MBG, the angles KAD, MBG are equal, as also the right angles AKD, BMG, and that the sides AK, BM, adjacent to the equal angles, are equal to one another j therefore KD is equal'' to AIG, and AD to BG: For the same reason, in the triangles KAL, MBN, KL is equal to MN, and AL to BN : And in the triangles LAD, NBG, LA, AD are equal to NB, BG, and they con- tain equal angles: therefore the base LD is equal' to the '4.1: base NG. Lastly, in the triangles KLD, MNG, the sides DK, KL, are equal to GM, MN, and the base LD to the base NG ; therefore the angle DKL is equal to"* the angle * 8. i. GMN : But the angle DKL is the inclination of the plane CAD to the plane CAE, and the angle GMN is the inclina- P 3 tion »6.de£.ll. 26. 1. 214 ' ' *^HZ ELEMENTS Book XI. tion of the, plane FBG to the plane FBH, which planes have »7*dcfn therefore the same inclination^ to one another: And in the ' sanme manner it may be demonstrated, that the other planes in w;hich the equal angles are, have the same inclination to one another. Therefore, if two solid angles, &c. Q^ E. D. PROP. B. THEOR. seeN. If two soUd aiiglcs bc contained, each by three plane angles which are equal to one another, each to each, and alike situated ; these solid angles are equal to one another. Let there be two solid angles at A and B, of which the solid angle at A is contained by the three plane angles CAD, CAE, EAD : and that at B, by the three plane angles FBG, FBH, HBG; of which CAD is equal to FBG; CAE to FBH; and EAD to HBG : The solid angle at A is equal to the solid angle at B. Let the solid angle at A be af)plied to the solid angle at B ; and, first, the plane angle CAD being applied to the plane angle FBG, so as the point A may coincide with the point B, and the straight line AC with BFj then AD coincides with BG, because the angle CAD is equal to the angle FBG : And because the inclination of ^ the plane CAE to the plane ^'' CAD is equal => to the inclina- tion of the plane FBH to the plane FBG, the plane CAE coincides with the plane FBH, because the planes CAD, FBG coincide with one another : And because the straight lines AC, BF coincide, and that the angle CAE is equal to the angle FBH ; therefore AE coin- cides with BH, and AD coincides with BG ; wherefore the plane EAD coincides with the plane HBG : Therefore the solid angle A coincides with the solid an2;le B, and conse- » 8. A. 1. quently they are equal ^ to one another. Q. E. D. OF EUCLID. PROP. C. THEOR. OOLID figures contained by the same number ofseeN. equal and similar planes alike situated, and having none of their solid angles contained by more than three plane angles ; are equal and similar to ©ne ano- ther. Let AG, KQ^be two solid figures contained by the same number of similar and equal planes, alike situated, viz. let the plana AC be similar and equal to the plane KM, the plane AF to KP ; BG to LQ_; GD to QN -, DE to NO; and, lastly, FH similar and equal to PR : The solid figure AG is equal and similar to the solid figure KQ^ Because the solid angle at A is contained by the three plane angles BAD, B AE, EAD, which, by the hypothesis, are equal to the plane angles LKiN, LKO, OKN, which contain the solid angle at K, each to each j therefore the solid angle at A is equal'' to the solid angle at K: In the same manner, the :. . other solid angles of the figures are equal to one another. If, then, the solid figure AG be applied to the solid figure KQ_, first the plane 4 G \E \ ■ ^ \! .0 Q. 1\ ^ M K figure AC being applied to the plane figure KM ; the straight line AB coinciding DV [ 's;;' N with KL, the figure AC must -^- B coincide with the figure KM, because they are equal and similar : Therefore the straight lines AD, DC, CB coincide with KN, NM, ML, each with each ; and the points A, D, C, B, with the points K, N, M, L: And the solid angle at A coincides with* the solid angle at K ; wherefore the plane AF coincides with the plane KP, and the figure AF with the figure KP, because they are equal and similar to one another : Therefore the straight lines AE, EF, FB, coincide with KO, OP, PL; and the points EF with the points O, P. In the same manner, the figure AH coincides with the figure KR, and the straight line DH with NR, and the point H with the point R: And be- cause the solid angle at B is equal to the solid angle at L, it may be proved, in the same manner, that the figure BG coin- P 4 cide« 2l6 THE ELEMENTS Book XI. cides with the figure LQ, and the straight line CG with MQ, ^^''^"^ and the point G with the point Q: Since, therefore, all the planes and sides of the solid figure AG coincide with the planes and sides of the solid figure KQ, AG is equal and si- milar to KQ^: And, in the same manner, any other solid fi- gures whatever contained by the same number of equal and similar planes, alike situated, and having none of their solid an- gles contained by more than three plane angles, may be proved to be equal and similar to one another. Q^ E. D. PROP. XXIV. THEOR. SeeN. • 16. 11. "10. 11. •=4. 1. "3*. 1. XF a solid be containecl by six planes, two and t\v@ of which are parallel ; tlie opposite planes are simi- lar and equal parallelograriis. Let the solid CDGH be contained by the parallel planes AC, GF ; BG, CE ; FB, AE : Its opposite planes are si- milar and equal parallelograms. Because the two parallel planes BG, CE, are cut by the plane AC, their common sections AB, CD are parallel*. Again, because the two parallel planes BF, AE are cut by the plane AC, their common sections AD, BC are parallel^ : And AB is parallel to CD ; therefore AC is a parallelogram. In like manner, it may be proved that each of the figures CE, FG, GB, BF. AE is a parallelogram : Join AH, DF; and because AB is paral- lel to DC, and BH to CF ; the two straight lines AB, BH, which meet one another, are parallel to DC and CF, which meet one another, and are not in the same plane with the other two ; wherefore they con- tain equal angles'' ; the angle ABH is therefore equal to the angle DCF : And because AB, BH arc equal to DC, CF, and the angle ABH equal to the angle DCF ; therefore the base AH is equal "^ to the base DF, and the triangle ABH to the triangle DCF : And the parallelogram BG is double^ of the triangle ABH, and the parallelogram CE double of the triangle DCF ; therefore the parallelogram BG is equal and si. ilar to the parallelogram CE. In the same manner it may be proved, that the parallelogram AC is equal and similar to OF EUCLID. 217 to the parallelogram GF, and the parallelogram AE to BF. BookXI. Therefore, if a solid, &c. Q, E. D. ^-^-^ PROP. XXV. THfeOR. XF a solid parallelopiped be cut by a plane parallel to ^**^' two of its opposite planes ; it divides the wiiole into two solids, the base of one of which shall be to the base of the other, as the one solid is to the other. Let the solid parallelopiped ABCD be cut by the plane EV, which is parallel to the opposite planes AR, HD,and di- vides the whole into the two solids ABFV, EGCD'j as the base AEFY of the first is to the base EHCF of the other, so is the solid ABFV to the solid EGCD. Produce AH both ways, and talce any number of straight lines HM, MN, each equal to EH, and any number AK, KL, each equal to EA, and complete the parallelograms LO, KY, HQ,, MS, and the solids Li', KR, HU, MT: Then, because thestraightlines LK, KA,AE are all equal, the parallelograms X L Z ES H C lis. \Y O Y J) H ^^M-4 SN^T M_ \LAi iN C "0. s LO, KY,AF are equal*: And likewise the parallelograms KX,»*^- ^• KB, AG* ; as also ^ the parallelograms LZ, KP, AR, because " -•*• ^' they are opposite planes : For the same reason, the parallelo- grams EC, HQ^, MS, are equaP ; and the parallelograms HG, HI, IN, as also^ HD, MU, NT : Therefore three planes of the solid L P are equal and similar to th ree planes of the solid K R, as also to three planes of the solid AV : But the three planes opposite to these three are equal and similar ^ to them in the several solids, and none of their solid angles are contained by more than three plane angles : Therefore the three solids LP, KR, AV are equal'= to one another: For the same reason the « c. i! three solids ED, HU, MT are equal to one a:iother : There- fore 2l8 THE ELEMENTS Book XI. (q^q what multiple soever the base LF is of the base AF, the '^"^'"^''^ same multiple is the solid LV of the solid AV : For the same reason, whatever multiple the base NF is of the base HF, the same multiple is the solid NV of the solid ED : And if the base LF be equal to the base NF, the solid LV is equal^ to the solid N V ; and if the base LF be greater than the base NF, the solid LV is greater than the solid NV ; and if less, less : Since then there are four magnitudes, viz. the two bases AF, FH, ^ c. 11. * 5. def. 5. and the two solids AV, ED, and of the base AF and solid AV, the base LF and solid LV are any equimultiples whatever ; and of the base FH and solid ED, the base FN and solid N V are any equimultiples whatever ; and it has been proved, that if the base LF is greater than the base FN, the solid LV is greater than the solid NV i and if equal, equal; and if less, less. Therefore'' a« the base AF is to the base FH, so is the solid AV to the solid ED. Wherefore, if a solid, &c. Q^ E. D. PROP. XXVL THEOR. SeeN. « 11. 11. w23. 1. '12.11. J^T a given point in a given straight line, to make a solid angle equal to a given solid angle contained by three plane angles. Let AB be a given straight line, A a given point in it, and D a given solid angle contained by the three plane angles EDC, EDF, FDC : It is required to make at the point A in the straight line AB a solid angle equal to the solid angle D. In the straight lineDF take any point F, from which draw "FG perpendicular to the plane EDC, meeting that plane in G; join DG, and at the point A, in the straight line AB, make'' the angle BAL equal to the angle EDC, and in the plane BAL make the angle BAK equal to the angle EDG j then make AK equal to DG, and from the point K ercd"' KH at OF EUCLID. 219 at right angles to the plane BAL: and make KH equal to GF, Book XI. and join AH : Then the solid angle at A, which is contained ^-^^"^^ by the three plane angles BAL, BAH, HAL, is equal to the solid angle at D contained by the three plane angles EDC, EDF, FDC. Take the equal straight lines AB, DE, and join HB, KB, FE, GE : And because FG is perpendicular to the plane EDC, it makes right angles ^ with every straight line meeting it in <3.def. 11., that plane : Therefore each of the angles FGD, FGEisa right angle: For the same reason, HKA,H1CB are right angles: And because KA, AB areequal to GD, DE, each to each, and con- tain equal angles, therefore the base BK is equal' to th^ base ' "*• '• EG: And KH is equal to GF, and HKB, FGE are right angles, therefore HB is equal* to FE: Again, because AK, KH arc equal to DG, GF : and contain right angles, the base AH is equal to the base DF : and AB is equal to DE : therefore HA, AB, are equal to FD, DE, and the base HB is equal to the base FE, therefore, the angle BAH is J^ equal*^ to the angle ^ ^ fg , EDF : For the same reason, the angle HAL is equal to the angle FDC. Be- cause if AL and DC be made equal, and KL, HL, GC, FC be joined, since the whole angle BAL is equal to the whole EDC, and the parts of them'BAK,EDG are, by the construc- tion, equal : therefore the remaining angle KAL is equal to the remaining angle GDC : And because KA, AL are equal to GD, DC, and contain equal angles, the base KL is equal* to the base GC : And KH is equal to GF, so that LK, KH are equal to CG, GF, and they contain right angles; therefore the base HL is equal to the base FC : Again, because HA, AL are equal to FD, DC, and thi.- base HL to the base FC, the angle HAL is equal' to the angle FDC: Therefore, because the three plane angles BAL, BAH, HAL, which contain the solid angle at A, are equal to the three plane angles EDC, EDF, FDC, v/hich contain the solid angle at D, each to each, and are situated in the same order, the s jlid angle at A is equals to the solid angle at D. Therefore, at a given point, in a jiven straight line, a solid angle has been made equal to a given solid angle contained by three plane angles. Which was to be done. THE ELEMENTS ^PRdP. XXVII. PROB. »26. 11. ^ 12. 6. "24.11. X O describe from a parallelepiped similar, given. en straight line k solid and similarly situated to one Let AB be the given straight line, and CD the given solid parallelepiped, .'^t is required from AB to describe a solid paralleiopiped similar, and similarly situated to CD. At the point A of the given straight line AB, make*, a solid angle equal to the solid angle at C, and let BAK, KAH, HAB be the three plane angles which contain it, so that BAK be equal to the angle ECG, and KAH to GCF, and HAB to FCE : And as EC to CG, so make^ BA to AK ; and as GC to CF, so make** KA to AH ; wherefore, ex oequali*^, as EC to CF, so is BA to AH : Complete the parallelogram BH, and the solid AL : And because, as EC to CG, so BA to AK, the sides about the equal angles ECG, BAK are propor- tionals ; therefore the parallelogram BK is similar to EG. For the same reason, the paral- lelogram KH is similar to GF, and HB to FE. Wherefore three parallelograms of the solid AL are similar to three of the solid CD; and the three opposite ones in each solid are equal"* and similar to these, each to each. Also, because the plane angles which contain ths solid angles of the figures are equal, each to each, and situated in the same order, the solid angles '^H def 11 ^^^ equal=, each to each. Therefore the solid AL is similar*' to the solid CD. Wherefore from a given straight line AB a solid paralleiopiped AL has been described similar, and simi- larly situated to the given one CD. Which was to be done. Fz «^1 D c E QF EUCLID, PROP. XXVIII. THEOR, IF a solid parallelopiped be cut by a plane passing see n. through the diagonals of two of the opposite planes; it shall be cut in two equal parts. Let AB be a solid paraIlelopip,ed, and DE, CF the diago- nals of the opposite parallelograms AH, GB, viz, those which are drawn betwixt the equal angles in each : And because CD, FE are each of them parallel to GA, and not in the same plane with itj CD, FE are parallel* ; wherefore the diagonals CF, *^- ^J- DE are in the plane in which the pa- rallels are, and are themselves paral- lels*': And the plane CDEF shall cut the solid AB into two equal parts. Because the triangle CGF is equaP to the triangle CBF, and the triangle DAE to DHEj and that the paral- lelogram CA is equal"" and similar to the opposite one BE j and the paral- lelogram GE to CH -, Therefore the prism contained by the tv.'o triangles CGF, DAE, and the three parallelograms CA, GE, EC, is equal'^to the prism contained by the two triangles CBF, DHE, ' C. ii and the three parallelograms BE, CH, EC ; because they are contained By the same number of equal and similar planes, alike situated, and none of their solid angles are contained by more than three plane angles. Therefore the solid AB is cut into two equal parts by the plane CDEF. Q^ E. D. *N. B. The insisting straight lines of a parallelopiped, men- * tioned in the next and some following propositions, are the ^ sides of the parallelograms betwixt the base and the opposite * plane parallel to it.' 16. 11^ «^34. 1. ^2i. 11. PROP. XXIX. THEOR. oOLID parallelopipeds upon the same base, and of See x. the same altitude, the insisting straight lines of which are terminated in the same straight lines in the plane opposite to the base, are equal to one another. Let k 222 Book XI. »28. 11. »34. 1. '38. 1. * 36. 1. e24. 11. THE ELEMENTS Let the solid parallelopipeds AH, AK be upon the same base AB, and of the same altitude, and let their insisting straight lines AF, AG, LM, LN be terminated in the same straight line FN, and CD, CE, BH, BK, be terminated in the same straight line DK ; the solid AH is equal to the solid AK, First, Let the parallelograms DG, HN, which are opposite to the base AB, have a common sideHG : Then, because the solid AH is cut by the plane AGHC passing through the diago- nals AG, CH of the opposite planes ALGF, CBHD, AH is cut into two equal parts* by the plane AGHC : Therefore the solid AH is double of the prism which is contained betwixt the triangles ALG, CBH: For the same reason, because the solid AK is cut by the plane I.GHB through the diagonals LG, BH of the opposite planes ALNG, CBKH, the solid AK is double of the same prism which is contained betwixt the triangles ALG, CBH, Therefore the solid AH is equal to the solid AK. But, let the parallelo<rrams DM, EN, opposite to the base, have no common side: Then, because CH, CK are parallelo- grams, CB is equaP to each of the opposite sides iJU, EK j wherefore DH is equal to EK: A.dd, or take away the common part HE ; then DE is tqual to HK : Wherefore also the tri- angle CDE is equal^ to the triangle BHK : And the parallelo- gram DG is equal* to the parallelogram HN : For the same reason, the triangle AFG is equal to the triangle LMN, and the parallelogram CF js equal* to the parallelogram BM, and CG to 5N : for they arc opposite. Therefore the prism which is contained by the two triangles AFG, CDE, and the three ' C. 11. parallelograms AD, DG, GC is equaU to the prism, contain- ed by the two triangles LMN, BHK, and the three parallelo- grams BM, MK, KL. If therefore the prism LMN B 11 K be taken OF EUCLID. 223 taken from the solid of which the base is the parallelogram Book XI. AB, and in which FDKN is the one opposite to it ; and if '"■^^"^-^ from this same solid there be taken the prism AFGCDE, the remaining solid, viz. the parallelepiped AH, is equal to the remaining parallelepiped AK. Therefore solid parallelo- pipeds, &c. Q. E. D, PROP. XXX. THEOR. OOLID parallelopipeds upon the same base, andSecN. of the same altitude, the insisting straight lines of which are not terminated in the same straight lines in theplaneoppositetothebase, areequaltooneanother. Let the parallelopipeds CM, CN, be upon the same base AB, and of the same altitude, but their insisting straight lines AF, AG, LM, LN, CD, CE, BH, BK, not terminated in the same straight lines : The solids, CM, CN are equal to one another. Produce FD, MH, and NG, KE, and let them meet one another in the points O, P, Q,, R ; and join AO, LP, BQ., CR} And because the plane LBHM is parallel to the opposite N K plane ACDF, and that the plane LBHM is that in which are the parallels LB, MFlPQ^, in which also is the figure BLPQ ; and the plane ACDF is that in which are the parallels AC, FDOR, in which also is the figure CAOR ; therefore the figures BLFC^, CAOR are in parallel planes : In like manner, because the plane ALNG is parallelto the oppositeplaneCBKE, and that the plane ALNG is that iii which are the parallels AL, 224 THE ELEMENTS Book XI. AL, OPGN, in which also is the figure ALPO ; and the plane '""^'^'^^ CBKEisthat in which are the parallels CB,RQEK, in which also h the figure CBQR j therefore the figures ALPO, CBQil are in parallel planes : And the planes ACBL, ORQP arc parallel j therefore the solid CP is a parallelepiped : But the solid Civl, of which the base is ACBL, to which FDHM is the » 2«. 11, opposite parallelogram, is equal* to the solid CP, of which the base is the parallelogram ACBL, to which ORQP is the one opposite; because they are upon the same base, and their insist- ing straight lines AF, AO, CD, CR ; LM, LP, BH, BQ are in the same straight lines FR, MQ : And the solid CP is equal^ to the solid CN : for they are upon the same base ACBL, and their insisting straight lines AO, AG, LP, LN; CR, CE, BQ^, BK are in the same straight lines ON, RK : Therefore the solid CM is equal to the solid CN. Wherefore solid parallelopipeds, &c. Q^ E. D. SeeN, PROP. XXXL THEOR. ►^ O LID parallelopipeds, which are upon equal bases, and of the same altitude, are equal to one another. Let the solid parallelopipeds AE, CF, be upon equal bases AB, CD, and be of the same altitude j the solid AE is equal to the solid CF. First, Let the insisting straight lines be at right angles to the bases AB, CD, and let the bases be placed in the same plane, and OF EUCLID. 225 and so as that the sides CL, LB be in a straight lincj there- ^oo^ xi, fore the straight line LM, which is at right angles to the plane ^"^^-^^^ in which the bases are, in the point L, ^s common* to the two 1 13^ u^ solids AE, CF ; let the other insisting lines of the solids b e AG, HK, BE ; DF, OP, CN : And first, let the angle ALB b e equal to the angle CLD ; then AL, LD ?.;<? in a straight line . » 14. 1. Produce OD, HB, and let then: meet in Q,^nd complete the solid paralleiopiped LR, the base of which is the parallelogram lyQ, and of which LM is one of its insisting straight lines': Therefore, because the parallelogram AB is equal to CD, as the base AB is to the base LQ, so is ■= the base CD to the same c 7 5^ LQ : And becaust: the selid paralleiopiped AR is cut by the plane LMEB, which is parallel to the opposite planes AK,DRj as the base AB is to the base LQ; so is^ the solid AE to the <" 25. n. solid LR : For the same reason, because the solid parallopiped CR is cut bv the plane LMFD, which is parallel to the opposite planes CP,'BR j as the base CD to the P F R base LQ, so is the \^K solid CF to the solid LR : But as the base AB to the base LQ, so the base CD to the base LQ, as before was proved ; Therefore as the solid AEto the solid LRj.so is the solid CF to the solid LR j and therefore the solid AE is equah to the solid CF. * 9. 5. But let the solid parallelopipedsSE, CF be upon equal bases SB, CD, and be of the same altitude, and let their insisting straight lines be at right angles to the bases ; and place the bases SB, CD in the same plane,so that CL, LB be in a straight line ; and let the angles SLB, CLD be unequal ; the solid SE is also in this case equal to the solid CF : Produce DL, TS until they meet in A,and from B draw BH parallel to DA ; and let HB, OD produced meet in Q, and complete the solids AE, LR: Therefore the solid AE, of which the base is the parallelo- gram LE, and AK the one opposite to it, is equal *" to the solid f 29. n. SE, of which the base is LE, and to which SX is opposite; for they are upon the same base LE, and of the same altitude, and their insisting straight lines, viz. LA, LS, BH, BT ; MG, MV, EK, EX are in the same straight lines AT, GX : And Q because 126 THE ELEMENTS Book XI. because the parallelogram AB is equals to SB, for they are upon tS5^^7r '^^ S^tne base LB, and between the same parallels LB, AT : and that the base SB is equal to the base CD ; therefore the base A B is equal to the base CD, and the angle ALB is equal to the angle CLD : Therefore, by the first case, the solid AE is equal to the solid CF ; but the solid AE is equal to the solid SE, as was demon- strated ; therefore the solid SE is equal to the solid CF. But, if the insisting straight lines AG, HK, BE,LM; CN, RS, DF, OP, be not at right angles to the bases AB, CDj in this case likewise the solid AE is equal to the solid CF : From the points G, K, E, M j N, S, F, P, draw the straight lines Ml. 11. GQ,KT,EV,MX;NY,SZ,FI, PU, perpendicular^ to the plane in which are the bases AB, CD ; and let them meet it in the points Q, T, V, X ; Y, Z, I, U, and join QT, TV, VX, XQ; YZ,ZI,ItJ,UY: Then, because GQKT are at right A S A H Q 1 "«. 11. angles to the same plane, they are parallel » to one another : And MG, EK are parallels ; therefore the plane MQ, ET, of which one passes through MG, GQ, and the other through EK, KT, which are parallel to MG, GQ, and not in the same "15. II. plane with them, are parallel"* to one another: For the same reason, the planes M V,GT are parallel toone another : There- fore the solid QE is a parallelepiped : In like manner, it may be proved, that the solid YF is a parallelopipcd : But, from what has been demonstrated, the solid EQ is equal to the solid FY, because they are upon equal bases MK, PS, and of the samealtitude,andhavetheirinsistingstraightlinesatrightangles 3 '• OF EUCLID. 127 to the bases : And the solid EQ is equal ' to the solid AE ; and ^^^ xi. the solid FY to the solid CF; because they are upon the same i^^^^ bases and of the same altitude: Therefore the solid AE is equal 11. to the solid CF : Wherefore solid parallelepipeds, &c. Q. E. D. " PROP. XXXII. THEOR. ♦Solid parallelopipeds which have tlie same alti- s«eN, tude, are to one another as their bases. Let AB, CD be solid parallelopipeds of the same altitude : They are to one another as their bases j that is, as the base AE to the base CF, so is the solid AB to the solid CD. To the straight line FG apply the parallelogram FH equal^ » c«r.45 1 to AE, so that the angle FGH be equal to the angle LCG ; and complete the solid parallelepiped GK upon the base FH, one of whose insisting lines is FD, whereby the solids CD,GK must be of the same altitude : Therefore the solid AB is equal*" * 31. 11. to the solid B N GK, because they are upon equal bases x/ » AE, FH, and ^r MO are of the same altitude: And because the so- lid parallelopi- A. K ped CK is cut by the plane DG which is parallel to its opposite planes, the base HF is = to the base FC, as the solid HD to the solid DC : c 23. 11. But the base HF is equal to the base AE, and the solid GK to the solid AB : Therefore, as the base AE to the base CF, so is the solid x\B to the solid CD. Wherefore solid parallelo- pipeds. Sec. Q. E. D. « CoR. From this it is manifest, that prisms upon triangular bases, of the same altitude, are to one another as their bases. Let the prisms, the bases of which are the triangles AEM, CFG, and NBO, PDQ the triangles opposite to them, have the same altitude ; and complete the parallelograms, A E,CF, and the solid parallelopipeds AB, CD,- in the first of which let MO, and in the other let GQ be one of the insisting lines. And because the solid parallelopipeds AB, CD have the same alti- tude, they are to one another as the base AE is to the base ' Q 2 CF ; <J 28. 11. •24. 11. "C. 11. M.G, THE ELEMENTS CF ; wherefore the prisms, which are their halves'*, are to one another, as the base AE to' the base CF j that is, as the triangle AEM to the triangle CFG. PROP. XXXIII. THEOR. oIAIILAR solid parallelopipeds are one to another in the triplicate ratio of their homologous sides. Let AB, CD be similar solid parallelopipeds, and the side AE homologous to the side CF : The solid AB has to the solid CD, the triplicate ratio of that which AE has to CF. Produce AE,GE, HE, and in these produced take EK equal to CF, EL equal to FN, and EM equal to FR ; and complete the parallelogram KL,and the solid KO : Because KE, EL, are equal to CF, FN, and the angle KEL equal to the angle CFN, because it is equal to the angle AEG, which is equal to CFN, by reason that the solids AB^CD are similar; therefore the parallelogram KL is similar and equal to the parallelogram CN. For the same reason, the parallelogram MK is similar and equal to CR, and also OEtoFD. There- fore three paralle- < J) lograms of the \ N. sol id KO are equal and similar to three parallelograms of \ the sol id CD: And the three opposite ones in each solid are equal » and similar to these : 7 herefore the so- O fid KO is equal'' and similar to the solid CD: Complete the parallelogram GK, and complete the solids EX, LP upon the bases GK, KL, so that EH be an insisting straight line in each of them, whereby they must be of the same altitude with the solid AB : And because the solids AB, CD are similar, and, by permu- tation, as AE is to CF, so is EG to FN, and so is EH to FR ; and B^C is equal to FK, and FN to EL, and FR to EM : Therefore, as AP^ to EK, so is EG to EL, and so is HE to EM: But, as AE to EK, so"^ is the parallelogram AG to the parallelogram GK ; and as GE to EL, so is " GK to KL, and N ^R c :f B X \ \V i\ P G \ \ K \ \ J ■ML \ U \l \ OF EUCLID. - 229 and as HE to EM, 50"= is PE to KM : therefore as the par?.]- ^°^^^- lelogram AG to the paralle]o:jram GK, so is GK to KL, and c j g PE to KM : But as AG, to GK, so ^ is the solid AB to the «« 25. ii. solid EX ; and as GK to KL, so^ is the solid EX to the so- lid PL ; and as PE to KM, so-i is the solid PL to the solid KO : And therefore as the solid AB to the solid EX, so is EX to PL, and PL to KO : But if four magnitudes be continual pro- portionals, the first is said to have to the fourth the triplicate ratio of that which it has to the second : Therefore the solid AB has to the solid KO, the triplicate ratio of that vhich AB has to EX : But as AB is to EX, so is the parallelogram AG to the parallelogram GK, and the straight line AE to the straight line EK. Wherefore the solid AB has to the solid KO, the triplicate ratioof that which AE has to EK. And the solid KO is equal to the solid CD, and the straight line EK isequal to the straight line CF. Therefore the solid AB has to the solid CD, the triplicate ratio of that which the side AE has to the homologous side CF, kc. Q^ E. D. Cor. From this it is manifest, that, if four straight lines be continual proportionals, as the first is to the fourth, so is the solid parallelopiped described from the first to the similar solid similarly described from the second ; because the first straight line has to the fourth the triplicate ratio of that which it has to the second. PROP. D. THEOR. OOLID parallelopipeds contained by parallelo- see n. grams equiangular to one another, each to each, that is, of which the solid angles are equal, each to each, have to one another the ratio M'hich is the same with the ratio compounded of the ratios of their sides. Let AB, CD be solid parallelopipeds, of which ABJs con- tained by the parallelograms AE, AF, AG equiangular, each to each, to the parallelograms CH, CK, CL, which contains the solid CD. The ratio which the solid AB has to the solid CD, is the same with that which is compounded of the ratios of the sides AM to DL, AN to DK, and AO to DH. Q_3 Produce 23® Book XI, •C. 11. *32. 11. THE ELEMENTS Produce MA, NA, OA to P, Q^, R, so that AP be equal to DL, AQ to DK, and AR to DH ; and complete the solid parallelopiped AX contained by the parallelograms AS, AT, AV similar and equal to CH, CK, CL, each to each. There- fore the solid AX is equal* to the solid CD. Complete likewise the solid AY, the base of which is AS, and of which AO is one of its insisting straight lines. Take any straight line a, and as MA to AP, so make a to b ; and as NA to AQj so make b to c ; and as AO to AR, so c to d : Then, because the parallelogram AE is equiangular to AS, AE is to AS, as the straight line a to C, as is demonstrated in the 23d Prop. Book 6, and the solids AB, AY, being betwixt the parallel planes BOY, EAS, are of the same altitude. Therefore the solid AB is to the solid AY, as"* the base AE to the base AS : that is, as the straight line a is to c. And the solid AY is to the solid D H L \ \ K \ \ a — •b — c — - d— C tNl AX, as' the base OQ^is to the baseQR ; that is, as the straight ■^ 25. 11. line OA to AR ; that is, as the straight line c to the straight line d. And because the solid AB is to the solid AY, as a is to c, and the solid AY to the solid AX, as c is to d } ex acquali, the solid AB is to the solid AX or CD which is equal t» it, as the straight line a is to d. But the ratio of a to d is said to *Def. A. 5. be compounded'^ of the ratios of a to b, b to c, and c to d, which are the same with the ratios of the sides MA to AP, N A to A(^, and OA to AR, each to each. And the sides AP, AQ_, AR are equal to the sides DL, DK, DH, each to each. Therefore the solid AB has to the solid CD the ratio, which is the same with that which is compounded of the ratios of the sides AM to DL, AN to DK, and AQ to DH. Q. E. D. OF EUCLID 231 Book XI. PROP. XXXIV. THEOR. "-^ The bases and altitudes of equal solid parallelo- seeN. piped, are reciprocally proportional; and if the bases and altitudes be reciprocally proportional, the solid parallelopipeds are equal. Let AB, CD be equal solid parallelopipeds ; their bases are reciprocally proportional to their altitudes; that is, as the base EH is to the base NP, so is the altitude of the solid CD to the altitude of the solid AB. First, Let the insisting straight lines AG, EF, LB, HK ; CM, NX, OD, PR be at right angles to the bases. As the base EH to the base NP, so is CM to AG. If the base EH be equal to the base NP, then because the solid AB is likewise equal to the solid CD, CM shall beaqualtoAG. Because if the bases EH, NP, be equal, but the altitudes AG, CM be not equal, neither shall the solid AB be equal to the solid CD. But the solids are equal, by the hypothesis. Therefore the altitude CM is not unequal to the altitude AG ; that is, they are equal. Wherefore, as the base EH to the base NP, so is CM to AG. Next, let the bases EH, NP not be equal, but EH greater than the other : Since then the solid AB is equal to the solid CD, CM is therefore greater than AG : For if it be not, neither also in this case would the solids AB, CD be equal, which, by the hypothesis, are equal. Make then CT equal to AG, and complete the solid parallelopiped CV of which the base is NP, and altitude R_D K B H \ G \1 L M A I\x -^ O C N CT. Because the solid AB is equal to the solid CD, therefore the solid AB is to the 0^4 solid 232 THE ELEMENTS Book XI. solid CV, as* the solid CD to the solid CV. But as the solid AB to the solid CV, so"* is the base Eri to the D^se NP ; for the solids AB, CV are of the same altitude ; and as the solid CD to CV, so= is the base MP to the base PT, and s./ is the straight line MC to CT ; and CT is crquai to AG. Therefore, as the base EH to tht base NP, so is MC to AG. Wherefore the bases of the- solid parallelopipeds AB, CD are reciprocally proportional to their altitudes. Let now the bases of the solid paiallelopipeds AB, CD be reciprocally proportional to their altitudes j viz. as the base EH to the base NP, so the al- titude of the solid CD to the altitude of the solid AB J the solid AB is equal to the solid CD. Let the insisting lines be, as before, at right angles to H the bases. Then, if the base EH be equal to the base NP, since EH is to NP, as the altitude of the solid CD is to the altitude of the solid AB, therefore the altitude of CD is equal'^ to the altitude of AB. But solid parallelopipeds upen equal bases, and of the same altitude, are equal ^ to one another j therefore the solid AB is equal to the solid CD. But let the bases EH, NP be unequal, and Jet EH be the greater of the two. Therefore, since as the base EH to the base; NP, so is CM thealti- K B 1 PL V ► \ (r \ "5? \M \- T, P \ \ \ \ 1 \ t ( * N sA. 5. '31. 11. tude of the solid CD to AG the al titude of AB, CM is greater'' than AG. Again, take CT equal to AG, and com- plete, as before, the so lid CV. And because the base EH is to the baseNP,asCMtoAG and that AG is equal to CT, therefore the base R D K I^ H \ \ M K N T \ \ A E C N X EH is to the base NP, as MC to CT. But as the base EH is to NP, so^ is the solid AB to the solid CV ; for the solids A3> CV are of the same altitude j and as MC to CT, so i$ the OF EUCLID. 233 the base MP to the base PT, and the solid CD to the solid' ^^^V^J* CV : And therefore as the solid AB to the solid CV, so is the c s^^^nT solid CD to the solid CV ; that is, each of the solids AB, CD has the same ratio to the solid C V : and therefore tha solid AB is equal to the solid CD. Second general case. Let the insisting straight lines FE, BL, GA, Kf^ ; XN, DO, iVlC, RP not be at right angles to the bases of rhe solids ; and from the points F, B, K,'G ; X, D, R, M draw perpendiculars to the planes in which are the bases EH, NP meeting those planes in the points S, Y, V, T; Q, 1, U, Z ; and complete the solids FV, XU, which are parallelnpipeds, as was proved in the last part of Prop. 31 of this Book. In this case, lilcewise, if the solids AB, CD be equal, their bases are reciprocally proportional to their altitudes, viz. the base EH to the base NP, as the altitude of the solid CD to the altitude of the solid AB. Because the solid AB is equal to the solid CD, and that the solid BT is equal? to the e'29or30. solid BA, for they are upon the same base FK, and of the ^^' same altitude; and that the solid DC is equals to the solid DZ, being upon the same base XR, and of the same altitude ; therefore the solid BT is equal to the solid DZ : But the bases are reciprocally proportional to the c 'titudes of equal solid parallelepipeds of r/hich the insisting straight lines are at right angles to their bases, as before was proved : Therefore as the base FK to the base XR, so is the altitude of the solid DZ to the altitude of the solid BT : And the base FK is equal to the base EH, and the base XR to the base NP : Wherefore, as the base EH to the base NP, so is the altitude of the solid DZ to the altitude of the solid BT : But the altitudes of the solids DZ, DC, as also of the solids BT, BA are the same. There, fore as the b^se EH to the br;se NP, so is the altitude of the solid »34 THE ELEMENTS Book XI. solid CD to the altitude of the solid AB ; that is, the bases of '"^^'^^'^^ the solid parallelopipeds AB, CD are reciprocally proportional to their altitudes. Next, Let the bases of the solids AB, CD be reciprocally proportional to their altitudes, viz. the base EH to the base NP, as the altitude of the solid CD to the altitude of the solid ABj the solid AB is equal to the solid CD: The sanae construction being made ; because, as the base EH to the base NP, so is the altitude of the solid CD to the altitude of the solid AB ; and that the base EH is equal to the base FK; and NP to XR ; therefore the base FK is to the base XR, as the altitude of the solid CD to the altitude of AB : But the alti- tudes of the solids AB, BT are the same, as also of CD and DZ i therefore as the base FK to the base XR, so is the alti- tude of the solid DZ to the altitude of the solid BT : Where- fore the bases of the solids BT, DZ are reciprocally propor- tional to their altitudes: and their insisting straight lines are at right angles to the bases ; wherefore, as was before proved, • 29 or 30. the solid BT is equal to the solid DZ; But BT is equals to "• the solid BA, and DZ to the solid DC, because they are upon the same bases, and of the same altitude. Therefore the solid AB is equal to the solid CD. Q^ E. D. OF EUCLID. PROP. XXXV. THEOR. 235 Book XI. XF, from the vertices of two equal plane angles, there See n. be drawn two straight lines elevated above the planes in which the angles are, and containing equal angles with the sides of those angles, each to each ; and if in the lines above the planes there be taken any points, and from them perpendiculars be drawn to the planes in which the first named angles are : And from the points in which they meet the planes, straight lines be drawn to the vertices of the angles first named ; these straight lines shall contain equal angles with the straight lines whicli are above the planes of the angles. Let BAC, EDF be two equal plane angles ; and from the points A, D let the straight lines AG, DM be elevated above the planes of the angles making equal angles with their sides, each to each, viz. the angle GAB equal to the angle MDE, and GAC to MDF ; and in AG, DM let any points G, M be taken, and from them let perpendiculars GL, MN be drawn to theplanes BAC, EDF, meeting these planes in the pointy L, Nj U and join LA, ND : The angle' GAL is equal to the angle Make AH equal to DM, and through H draw HK parallel to GL: But GL is perpendicular to the plane BAC; where- fore HK is perpendicular" to the same plane : From the points K, N, to the straight lines AB, AC, DE, DF, draw perpen- diculars KB, KC, N£, NFj and join HB, BC, ME, EF : Because 8. 11. «4.clef. 11, 'S.def.ll, e 26. 1. THE ELEMENTS Because HK is perpendicular to the plane BAC, the plane HBK v/hich passes through HK is at rightangles*' to the plane BAC ; and AB is drawn in the plane BAC at right angles to the con^mon section BK of the two planes ; therefore AB is perpendicular'^ to the plane HBK, and makes right angles'" with every straight line meetingitin that plane: But BH meets it in that plane ; therefore ABH is a right angle : For the same reason, DEM is a right angle, and is therefore equal to the angle ABH : And the angle HAB is equal to the angle MDE. Therefore in the two triangles HAB, MDE there are two angles in one equal totwoangles in the other, each to each, and one side equal to one side, opposite to one of the equal angles in each, viz. HA equal to DM j therefore the remaining sides are equal<=, each to each : Wherefore AB is equal to DE. In the same manner, if HC and MF be joined, it may be demon- strated that AC is equal to DF : Therefore, since AB is equal to DE, BA and AC are equal to ED and DF j and the angle BAC is equal to the anle EDF j wherefore the base BC ■4. i; equaF to the base EF, and the remaining angles to the remain- ing angles : The angle ABC is therefore equal to the angle DEF : And the right angle ABK is equal to the right angle. DEN, whence the remaining angle CBK is equal to the remaining angle FEN : For the same reason, the angle BCK is equal to the angle EFN : Therefore in the two triangles BCK, EFN, there are two angles in one equal to two angles in the other, each to each, and one side equal to one side adjacent to the equal angles in each, viz. BC equal to EF ; the other sides, therefore, are equal to the other sides ; BK then is equal to EN: and AB is equal to DE ; wherefore AB, BK are equal to DE, EN ; and they contain right angles ; wherefore the base AK is equal to the br.sc DN : And since AH is equal to OF EUCLID. 23f DM, the square of AH Is equal to the square of DM: But the Book xr. squares of AK, KH are equal to the squares of AH, because g^^i AKH is a right angle : And the squares of DN, NM are equal to the square of DM, for DNM is a right angle: Wherefore the iquarcs of AK; KH are equal to the squares of DN, NM; and of those the square of AK is equal to the square of DN : Therefore the remainincr square of KH is equal to the remain- ing square of NM ; and the straight line KH to the straight line NM : and because HA, AK are equal to MD, DN, each to each, and the base HK to the base MN, as has been proved ; therefore the angle HAK is equal ^ to die angle MDN. '"S, i. Q, E. D. Cor. From this it is manifest, that if, from the vertices of two equal plane angles, there be elevated two equal straight lines containing equal angles with the sides of the angles, each to each j the perpendiculars drawn from the extremities of the equal straight lines to the planes of the first angles are equal to one another. Another Demonstration of the Corollary. Let the plane angks BAC, EDF be equal to one another, and let AH, DM, be two equal straight lines above the planes of the angles, containing equal angles with BA, AC ; ED DF, each to each, viz. the angle HAB, equal to MDE, and HAC equal to the angle MDF ; and from H, M let HK, MN be perpendiculars to the planes BAC, EDF : HK is equal to MN.. Because the solid angle at A is contained by the three plane angles BAC, BAH, HAC, which are, each to each, equal to the three plane angles EDF, EDM, MDF containing the solid angle at D ; the solid angles at A and D are equal, and there- fore coincide withone another ; to wit, ifthe plane angle BAC be applied to the plane angle EDF, the straight line AH coin- cides with DM, as was shewn in Prob. B. ol" this Book : And because AH is equal to DM, the point H coincides with the point M: Wherefore HK, which is perpendicular to the plane BAC, coincides with MN» which is perpendicular to the plane ' 13. u. EDF, because these planes coincides v/ith one another : Therefore HK is equal to MN. Q. E. D. • 238 THE ELEMENTS Book XI. SteN. •26.11. 14.6. PROP. XXXVI. THEOR. 1 F three straight lines be proportionals, the solid parallelopiped described from all three as its sides, is equal to the equilateral parallelopiped described from the mean proportional, one of the solid angles 1 of which is contained by three plane angles equal, \ each to each, to the three plane angles containing one of the solid angles of the other figure. Let A, B, C be three proportionals, viz. A to B, as B to C: . The solid described from A, B, C is equal to the equilateral solid described from B, equiangular to the other. ; Take a solid angle D contained by three plane angles EDF, FDG, GDE; and make each of the straight lines ED, DF, ' DG equal to B, and complete the solid parallelopiped DH : j Make LK equal to A, and at the point K in the straight linef LK, make ^ a solid angle contained by the three plane angles LKM, MKN, NKL equal to the angles EDF, FDG, GDE, H \ ' \ \ X M N F \ \ N \ G K E D B each to each; and make KN equal to B, and KM equal to C; and eomplete the solid parallelopiped KG : And because, as A is to B, so is B to C, and that A is equal to LK, and B to each of the straight lines DE, DF, and C to KM ; there- fore LK is to ED, as DF to KM ; that is, the sides about the equal angles are reciprocally proportional j therefore the pa- rallelogram LM is equal^ to EF; and because EDF, LKM are two equal plane angles, and the two cciual straight lines DG, KN are drawn from their vertices above their planes, and con- tain equal angles with their sides j therefore the perpendi- culars from the points G, N, to the planes EDF, LKM are equal OF EUCLID. 239 equal*^ to one another : Therefore the solids KO, DH are of ^^^J^^ the same altitude ; and they are upon equal bases LM, EF, c Co^ 33. and therefore they are equal'' to one another: But the solid n. KO is described from the three straight lines A, B, C, and *^^- *'* the solid DH from the straight line B. If therefore three straight lines, &c. Q^ E. D. PROP. XXXVII. THEOR. Xf four straight lines be proportionals, the similar s«3f- solid parallelopipeds similarly described from them shall also be proportionals. And if the similar pa- rallelopipeds similarly described from four straight lines be proportionals the straight lines shall be proportionals. Let the four straight lines AB, CD^ EF, GH be propor- tionals, viz. as AB to CD, so V.F to GH ; and let the simi- lar parallelopipeds AK, CL, EM, GN be similarly described from them. AK is to CL, as EM to GN. Make* AB, CD, O, P continual proportionals, as also EF, »n.6. GH, Q^ R : And because as AB is to CD, so EF to GH : and K B C2J 3 S d K that CD is*" to O, as GH toQ_, and O to P, as Q,, to R ; there- » n. 5. fore, ex aequali'', AB is to P, as EF to R : But as AB to P, c22. 5, SO"* is the solid AK, to the solid CL ; and as EF to R, so'' is " ^'"'■•33. the solid EM to the solid GN : Therefore'' as the solid AK to the solid CL, so is the solid EM to the solid GN. But 11. i4^ THE ELEMENTS *2Y. n. Book Xf. But let the solid AK be to the solid CL^ as the solid EM to the solid GN : The straight line AB ib to CD, as EFto GH. Take AB to CD, as Ev to ST, and from ST describe <= a solid parallelepiped SV similar and similarly situated to either of the solids £M, GN : And because AB is to CD, is EF to ST, and that from AB, CD the solid parallelopipeds AK, CL are similarly described j and in like manner the solids EM, SV from the straight lines EF, ST 3 therefore AK is to CL, as i. *9.5. SccN, Kl ^ N 1 \ 1 \ ■ J I B K 1) (; H d R EM to SV : But, by the hypothesis, AK is to CL, as EM to GN : Therefore GN is equaK to SV : But it is like'A^ise similar and similarly situated to S V; therefore the planes which contain the solids GN, SV are similar and equal, and their hbmologous sides GH, ST equal to one another : iVnd because as AB to CD, so EF to ST, and that ST is equal to GH: AB is to CD, as EF to GH. Therefore, if four straight lines, &c, Q. E. D, PROP. XXXVHL THEOR. If a plane be perpendicular to another plane, and " a straight line be drawn from a point in one of " the planes perpendicular to the other plane, this *' straight line shall fall on the common section of " the planes." " Let the plane CD be perpendicular to the plane AB, and *' let AD be the common se«Sion j if any point E be taken in " the piane CD, the perpendicular drawn from E to the plane « AB shall fall on AD. 3 " f Of OF EUCLID. 141 «3,def.lL « For, if it does not, let it, if possible, fall elsewhere, as EF ; ^^^;^- " and let it meet the plane AB in the point F j and from F « draw% in the plane AB a perpendicular FG to DA, which » 12. 1. •' is also perpendicular *" to the plane CD j and join EG : Then » 4. dcf. 11. " because FG is perpendicular •* to the plane CD, and the ** straight line EG, which is in •* that plane, meets it j there- ** fore FGE is a right angle'^ : ** But EF is also at right angles ** to the plane AB ; and there- *' fore EFG is a right angle : ** Wherefore two of the angles " of the triangle EFG are equal together to two right angles ; *' which is absurd : Therefore the perpendicular from the " point E to the plane AB, does not fall elsewhere than upon " the straight line AD; it therefore falls upon it. If there- « fore a plane," &c. Q^ E. D. PROP. XXXIX. THEOR. In a solid parallelopiped, if the sides of two of the SeeN. opposite planes be divided, each into two equal parts, the common section of the planes passing through the points of division, and the diameter of the solid parallelopiped cut each other into two equal parts. Let the sides of the opposite planes 11 K K CF, AH of the solid parallelopiped AF, be divided each into two equal parts in the points K, L, M, N; X, O, P, R; and join KL, MN, XO, PR; And be- cause DK, CL are equal and parallel, KL is parallel* to ^ H^^^-X I |^* XI "33. li. DC : For the same reason, MN is pa- rallel to BA : And •242 Book. XI. »5. 11. « 29. 1. « 4. 1. «U. 1. "33, 1. * 15. 1. 8 26. 1. THE ELEMENTS BA is parallel to DC j therefore, because KL, BA are each of them parallel to DC, and not in the same plane with it, KL is parallel'' to BA : And because KL, MN are each of them parallel'to BA, and not in the same plane with it, KL is parallel ^'to MN; wherefore KL, MN are in one plane. In like manner it may be proved, that XO, PR arte in one plane. Let YS be the common section of the planes KN, XR ; and DQ the diameter of the solid parailelopiped AF : YS and DG dd ipeet, and cut one another into two equal parts. Join DY, YE, BS, SG. Because DX is parallel to OE, the alternate angles DXY, YQE are equal"^ to one another : And because DX is equal to OE, and XY to YO, and contain equal an- gles, the base DY is equal ^ to the base YE, and the other angles are equal ; therefore the angle XYD is equal to the angle OYE, and Dye is a straight" * line : For the same reason BSG is a straight line, and BS equal to SG : And because CA is equal and parallel to DB, and also equal and parallel to EG; therefore DB is equal and parallel^ to EG : And DE, BG join their extremities ; therefore DE is equal and parallel * to BG : And DG, YS are drawn from po;nts in the one, to points in the other ; and are therefore in one plane : Whence it is manifest, that DG, YS must mtet one another ; let them meet in T : And because DE is pa- rallel to BG, the alternate angles EDT, BG I' are equal*^: and the angle DTY is equaK to the angle GTS : Therefore in the triangles DTY, GTS there are two angles in the one equal to two angles in the other, and one side equal to one side, opposite to two of the equal angles, viz. DY to GS ; for they are the halves ot DE, BG : Therefore the remaining sides are equaiE, each to each. Wherefore DT is equal to TG, aod YT equal to TS. Wherefore, if in a solid, 6ic. Q. E. D. OF EUCLID. PROP. XL. THEOR. XF there be two triangular prisms of the sanie al- titude, the base of one of which is a parallelogram, and the base of the other a triangle ; if the paral- lelogram be double of th« triangle, the prisms shall be equal to one another. Let the prisms ABCDEF, GHKLMN be of the same alti- tude, the first whereof is contained by the two triangles ABE, CDF, and the three parallelograms AD, DE, EC ; and the other by the two triangles GHK, LMN, and the three paral- lelograms LH, HN, NG J and let one of them have a paral- lelogram AF, and the other a triangle GHK for its base ; if the parallelogram AF be double of the triangle GHK, the prism ABCDEF is equal to the prism GHKLMN. Complete the solids AX, GO ; and because the parallelo- gram AF is double of the triangle GHK j and the parallelo- gram HK double* of the same triangle; therefore the paral- »34. i. lelogram AF is equal to HK. But solid parallelopipeds upon equal bases, and of the same altitude, are equaP to one another, b ^j^ jj Therefore the solid AX is equal to the solid GO j and the prism ABCDEF is halt* of the solid AX; and the prism <«« ,, GHKLMN half ^ of the solid GO. Therefore the prism * ABCDEF is equal to the prism GHKLMN. Wherefore, if there is two, &c. Q^ E. D. R ? [ 244 ] THE ELEMENTS or EUCLID. Book XII. BOOK XII. SeeN. LEMMA I. Which is the first preposition of the tenth book, and is neces- sary to some of the propositions of this book. J.F from the greater of two unequal magnitudes, there be taken more than its half, and from the re- mainder more than its half; and so on : There shall at length remain a magnitude less than the least of the proposed magnitudes. Let AB and C be two unequal magnitudes, of which AB is the greater. If from AB there be taken more that its half, and from the remainder more than its half, and so on ; there shall at length remain a magnitude less than C. For C may be multiplied so as at length to become greater than AB. Let it be so multi- plied, and let DE its multiple be greater than AB, and let DE be divided into DF, FG, GE, H each equal to C. From AB take BH greater p than its half, and from the remainder AH take HK greater than its half, and so on, until there be as many divisions in AB as there are in DE : And let the divisions in AB be AK, KH, HB ; and the divisions in ED be DF, FG, GE. And because DE is greater than AB, and A D K F \ B C E that THE ELEMENTS OF EUCLID. 245 that EG taken from DE is not greater than its half, but BH Book xii. taken from AB is greater than its half; therefore the remain- ^'^^''^'^ der GD is greater than the remainder HA. Again, because GD is greater than HA, and that GF is not greater than the half of GD, but HK is greater than the half of HA ; therefore the remainder FD is greater than the remainder AK. And FD is equal to C, tlierefore C is greater than AK j that is, AK is less than C. Q. E. D. And if only the halves be taken away, the same thing may in the same way be demonstrated. PROP. I. THEOR. oliVIILAR polygons inscribed in circles, arc to one another as the squares of their diameters. Let ABODE, FGHKL be two circles, and in them the similar polygons ABODE, FGHKL j and let BM, GN be the diameters of the circles: As the square of BM is to the square ofGN, so is the polygon ABODE to the polygon FGHKL. Join BE, AM, GL, F\ : And because the polygon ABODE is similar to the polygon FGHKL, and similar polygons are di- vided into similar triangles ; the triangles ABE , FGL, are similar and equiangular'' ; and therefore the angle AEB is equal to the * 6. 6. angle FLG: But AEB is equal"^ to AMB, because they stand *2i. 3. upon the same circumference ; and the angle FLG is, for the same reason, equal to the angle FNG : Therefore also the angle AMB is equal to FNG : And the right angle BAM is equal to the right** angle GFN ; wherefore the remaining an- * 31. 3, gjes in the triangles ABM, FGN are equal, and they are equi- R 3 angular 24^ THE ELEMENTS Book XII. angular to one another : Therefore as BM to GN,' so* is BA ^"^o"^"^ to GF ; and therefore the duplicate ratio of BM to GN, is f 10 def. 5. the same*^ with the duplicate ratio of BA, to GF : But the ratio «ML 6*' °^ ^^^ square of BM to the square of GN, is the duplicates ratio of that which BM has to GN j and the ratio of the po- lygon ABCDE to the polygon FGHKL is the duplicates of that which BA has to GF : Therefore as the square of BM to the square of GN, so is the polygon ABCDE to the poly- gon FGHKL. Wkerefore similar polygons, &c. Q. E. D. PROP. II. THEOR. Sfe N. V^IRCLES are to one another as the squares of their diameters. Let ABCD, EFGH be two circles, and BD, FH their dia- meters ; As the square of BD to the square of FH, so is the circle ABCD, to the circle EFGH. For, if it be not so, the square of BD shall be to the square of FH, as the circle ABCD is to some space either less than the circle EFGH, or greater than it*. First let it be to a space S less than the circle EFGH; and in the circle EFGH describe the square EFGH. This square is greater than half of the circle EFGH ; because if, through the points E, F, G, H, there be drawn tangents to the circle, the square • For tliere is some square equal to the rirck ABCD; let P be the wde of it, and to three straight lines BD, FH,and P, tlicre can l>ea fourlh jiroportional ; let this be Q : Therelbrc the squares of these four siraighl liiic« are proportionals; that is, to the »|uaresof BD, FH, and^the circle ABCB), it is possible there may be a fourth pro- portionaL Let this be S. Aud in like inaimcr are to be understood soanc things ii^ ioni« of the foUswing proposition*. OF EUCLID. 24-7 square EFGH.is half* of the square described about thecirclci BookXII. and the circle is less than the square described about it; there- .^^j fore the square EFGH is greater than half of the circle. Divide the circumferences EF, FG, GH, HE, each into two equal parts in the points, KjLjM,N,arui join EK,KF, FL,T..^, GM, MH, HX, NE : Therefore each of the triangles EKF, FLG, GMH, HNE is greater than half of the segment of the circle it stands in ; because, if straight lines touching the circle be. drawn through the points K, L, >I, N, and the parallelograms upon the straight lines EF, FG, GH, KE, b: completed ; each of the triangles EKF, FLG, GMH, HNE shall be the half^ of the parallslogram in which it is ; But every segment is less than the parallelogram in which it is : Wherefore each of the triangles EKF, FLG, GMH, HNE is greater than half the segment of the circle which contains it : And if these circum- ferences before named be divided each into two equal parts, and their extremities be joined by straight lines, by continuing to do this, there will at length remain segments of the circle, which, together, shall beless than the excessof thecircle EFGH, above the space S : Because, by the preceding Lemma, if from the greater of two unequal magnitudes there be taken more than its half, and from the remainder more than its half, and so on, there shall at length remain a magnitude less than the least of the proposed magnitudes. Let then the seg- ments EK, KF, FL, LG, GM, MH, HN, NE be those that remain, and are together less than the excess of the circle EFGH above S: Therefore the rest of the circle, viz. the polygon EKFLGMHN, is greater than the space S. Describe likevi'ise in the circle ABCD the polygon AXBOCPDR similar to the polygon EKFLGMHN : As therefore, the square of BD is to the square of VH^, so is the polygon AXBOCPDR co the i> I. '«?. polygon EKFLGMHN : But the square ofBD is also to the R 4 square 248 THE ELEMENTS Book XII. square of FH, as the circle ABCD is to the space S . There- j^JpC*"^ fore as the circle ABCD is to the space S, so is'= the polygon ' * AXBOCPDR to the polygon EKFLGMHN : But the circle ABCD is greater than the polygon contained in it ; wherefore * U.5, the space S isgreater'' than the polygon EKFLGMHN : But it is likewise less, as has been demonstrated; which is impos- sible. Therefore the square of BD is not to the square of FH, as the circle ABCD is to any space less than the circle EFGH; In the same manner, it may be demonstrated, that neither is the square of FH to the square of BD, as the circle EFGH is t» any space less than the circle ABCD. Nor is the square of BD to the square of FH, as the circle ABCD is to any space greater than the circle EFGH : For, if possible, let it be so to T,aspacegreaterthanthecircleEFGH: Therefore, inversely, as the square of FH to the square of BDt so is the space T to the circle ABCD. But as the spacef T is to the circle ABCD, so is the circle EFGH to some space, which must be less*" thaa the circle ABCD, because the space T is greater, by hypothe- sis, than the circle EFGH. Thereforeas thesquareofFH is to the + For, as in the foreeoing note at», it was explained how it was possible there could be a fourth proportional to the squares ot BD, FH, and the c.rcle ABCD, which was named S. So, in like manner, <^'"f,f»" ^'i! » J""'-'.*' fX^'"""' ^ this other space, named T, and the c.rcles ABCD, EFGH. And the hkc » to W understood ia some of the following pcooositioni. OF EUCLID. 249 the square of BD, so is the circle EFGH to a space less than ^^^^ '^^^' the circle ABCD, which has been demonstrated to be impos- "'^'^'^^'^ sible : Therefore the square of BD is not to the square of FH as the circle ABCD is to any space greater than the circle EFGH : And it has been demonstrated, that neither is the square of BD to the square of FH, as the circle ABCD to any space less than the circle EFGH : Wherefore, as the square of BD to the square of FH, so is the circle ABCD to the circle EFGHf . Circles therefore are, &c. Q^ E. D, PROP. III. THEOR. ti VER Y pyramid having a triangular base, may be see x. divided into two equal and similar pyramids having triangular bases, and which are similar to the whole pyramid ; and into two equal prisms wliich together are greater than half of the whole pyramid. Let there be a pyramid of which the base is the triangle ABC and its vertex the point D: The pyramid ABCD may be di- vided into two equal and similar pyra- mids having triangular bases, and simi- lar to the whole ; and into two equal prisms which together are greater than half of the whole pyramid. Divide AB, BC,'CA, AD, DB, DC, each into two equal parts in the; points E, F, G, H, K, L, and join EH, EG, GH, HK, KL, LH, KK, KF, FG. Because AE is equal to EB, and AH to HD, HE is parallel =• to DB : For the same reason, HK is parallel to AB; Therefore HEBK is a parallelogram, and HK equal ^ to EB: but EB is equal to AE ; therefore also AE is equal to HK : And AH is equal to HD; where- fore EA, AH are equal to KH, HD, each to each ; and the angle t AH is equal"^ to the angle KHD ; therefore the base EH is equal to the baseKD, and the triangle AEH " 34. 1. <=29. + ^fc»"se *»a fourth proportional I0 the square? cf BD, FH, and thf circl- ABC D, is possible, atwi that ilcaii-neitLcr be i^» iior greater than the circie iFGK, mast \m equal to it. 250 10. n. <"C. 11. «4. 6. "21. 6. »B. 11. & ll.dcf. 11 "41. 1. THE ELEMENTS AEH equal^ and similar to the triangle HKD : For the same reason, the triangle AGH is equal and similar to the triangle HLD : And because the two straight lines EH, HG, which meet one another, are parallel to KD, DL that meet one ano- ther, and arc not in the same plane WJth them, they contain equai^ angles; therefore the angle EHG is equal to the angle KDL. Again, because EH, HG are equal to KD, DL, each to each, and the angle EHG equal to the angle KDL ; there- fore the base EG is equal to the base KL : And the triangle EHG equaF and simi,lar to the triangle KDL : For the same reason, the triangle AEG is also equal and similar to the tri- angle HKL. Therefore the pyramid, of which the base is the triangle AEG, and of which the vertex is the point H, is equal ^ and similar to the pyramid the base of which is the triangle KHL, and vertex the point D : And because HK is parallel to AB, a side of the triangle ADB, the triangle ADB is equiangu- lar to the triangle HDK, and their sides are proportionals^ ; Therefore the triangle ADB is similar to the triangle HDK: And for the same reason, the triangle DBC is similar to the triangle DKL; and the "triangle ADC to the triangle HDL ; and also the triangle ABC to the triangle AEG: But the triangle AEG is similar to the triangle HKL, as before was proved ; therefore the triangle ABC is similar '^ to the triangle HKL. And the pyramid of which the base is the triangle ABC, and vertex the point D , is therefore S'imilar^ to the pyranjid of which the base is the tri- angle HKL, and vertex the same point D: But the pyramid of which the base is the triangle HKL, and vertex the point D, is similar, as has be n proved, to thepyramid the base of which is the triangle AEG, and vertex the point H : Wherefore the pyramid, the base of which is the triangle ABC, and vertex the point D, is similar to the pyramid of which the base is the tri- angle AEG and vertex H : Therefore each of the pyramids AEGH, HKLD is similar to the whole pyramid ABCD: And because BE is equal to EC, the parallelogram EBFG is double"^ of the triangle GFC : But when there are two prisms of the OF EUCLID. 251 same altitude, of which one has a parallelogram for its base, ^'^^ ^^• and the other a triangle that is half of the parailelogram, these prisms are equal^ to one another ; therefore the prism having ,^ 2jj. the parallelogram EBFG for its base, and the straight line KH opposite to it, is equal to the prism having the triangle GFC for its base, and the triangle HKL opposite to it; for they are of the same altitude, because they are between the parallel^ planes ABC, HKL : And it is manifest that each of * ^^- ''• these prisms is greater than either of the pyramids of which the triangles AEG, HKL are the bases, and the vertices the points H, D J because, if EF be joined, the prism having the parellelogram EBFG for its base, and KH the straight line opposite to it, is greater than the pyramid of which the base is the triangle EBF, and vertex the point K ; but this pyra- mid is equal*^ to the pyramid the base of which is the triangle '^•^*- AEG, and vertex the point H ; because they are contained by equal and similar planes : Wherefore the prism having the parallelogram EBFG for its base, and opposite side KH, is greater than the pyramid of which the base is the triangle AEG, and vertex the point K : And the prism of which the base is the parallelogram EBFG, and opposite side KH is equal to the prism having the triangle GFC for its base, and HKL the triangle opposite to it; and the pyramid of which the base is the triangle AEG, and vertex H, is equal to the pyramid of which the base is the triangle HKL, and vertex D : Therefore the two prisms before-mentioned are greater than the two pyramids of which the bases are the triangles AEG, L, and vertices the points H, D. Therefore the whole P^ amid of which the base is the triangle ABC; and vertex- the point D, is divided into two equal pyramids similar to one another, and to the whole pyramid ; and into two equal prisms ; and the two prisms are together greater than half of the whole pyramid. CL E. D. 252 THE ELEMENTS Book Xir. '"'^'^ PROP. IV. THEOR. secN. J^p there be two pyramids of the same altitude, upon triangular bases, and each of them be divided into two equal pyramids similar to the whole pyra-' mid, and also into two equal prisms ; and if each of these pyramids be divided in the same manner as the first two, and so on: As the base of one of the first two pyramids is to the base of the other, so shall all the prisms in one of them be to all the prisms in the other that are produced by the same number of divisions. Let there be two pyramids of the same altitude upon the tri- angular bases ABC, DEF, and having their vertices in the points G, H; and let each ot them be divided into two equal pyramids similar to the vfhole, rnd into two equal prisms; and let each of the pyramids thus made be conceived to be divided in the like manner, and so on: As the base ABC is to the base DEF, so are all the prisms in the pyramid ABCG to all the prisms in the pyramid DEFH made by the same number of divisions. Make the same construction as in the foregoing proposition ; And because BX isequal to XC,and AL to LC, therefore XL • '^- 6- is parallel'' to AB, and the triangle ABC similar to the tri- angle LXC : For the same reason, the triangle DEF is similar to RVF: And because BC is double of CX, and EF double of FV, therefore BC is to CX, as EF to FV: And upon BC, CX are described the similar and similarly situated redtilineal figures ABC, LXC ; and upon EF, FV, in like manner, are described the similar figures DEF, RVF : Therefore, as the * 22. 6. triangle ABC is to the triangle LXC, so'' is the triangle DEF to the triatigleRVF, and, by permutation, as the triangle ABC to the triangle DEF, so is the triangle LXC to the triangle R VF : And because the planes ABC, OMN, as also theplanes e 15. 11. DEF, STY are parallel, the perpendiculars drawn from the points G, H to the bases ABC, DEF, which, by the hypothe- sis, are equal to one another, shall be cut each into two equal " 17. n. "iparts by the planes OMN, STY, because the straight lines GC, HF are cut into two equal parts in the points N, Y by the same planes : Therefore theprismsLXCOMN, R VEST Y are of the same altitude j and therefore> as the base LXC to ,tlic OF EUCLID. ^53 ,32. the base RVF ; that is, as the triangle ABC to the triangle Book xii. DEF, so* is the prism having the triangle LXC for its base, ^ and OMN the triangle opposite to it, to the prism of which the ^ base is the triangle RVF, and the opposite triangle STY : ■ And because the two prisms in the pyramid ABCG are equal to one another, and also the two prisms in the pyramid DEFH equal to one another, as the prism of which the base is the parallelogram KBXL and opposite side MO, to the prism having the triangle LXC for its base, and OMN the triangle opposite to it ; so is the prism of which the base** is the paralle- * i. logram, PEV^R, and opposite side TS, to the prism of which the base is the triangle RVF, and opposite triangles STY. Therefore,componendo,as the prisms KBXLMO,LXCOMN together are unto the prism LXCOMN ; so are the prisms PEVRTS, RVFSTY to the prism R VFSTY : And permu- taado, as the prisms KBXLMO, LXCOMN are to the prisms PEVRTS, RVFSTY ; so is the prism LXCOMN to the prism RVFSTY : But as the prism LXCOMN to the prism RVFSTY, so is, as has been proved, the base ABC to the base DEF: Therefore, as the base ABC to the base DEF, so are the two prisms in the pyramid ABCG to the two prisms in the pyramid DEFH : And likewise if the pyramid? now made, for example, the two OMNG, ST YH be divided iri the same manner ; as the base OMN is to the base STY, so shall the two prisms in the pyramid OMNG be to the two prisms in the pyramid ST YH : But the base OiMN is to the base STY, as the base ABC to the base DEF ; therefore, as the base ABC to the base DEF, so are the two prisms in the pyramid ABCG to 254 THE ELEMENTS Book xii. ^q (he two prisms in the pyramid DEFH ; and so are thctw© ^ prisms in the pyramid OMNG to the two prisms in the py- ramid STYH ; and so are all four to all four: And the same thing may be shewn of the prisms made by dividing the pyra- mids AKLO and DPRS, and of all made by the same num- ber of divisions. Q. E. D. PROP. V. THEOR. JL VRAM JDS of the same altitude which have tri- angukr bases, are to one another as their bases. Let the pyramids of which the triangles ABC, DEF are the bases, and of which the vertices are the points G, H, be of the same altitude : As the base ABC to the base DEF, so is the pyramid ABCG to the pyramid DEFH. For, if it be not so, the base ABC must be to the base DEF, as the pyramid ABCG to a solid either less than the pyramid DEFH, or greater than it*. First, let it be to a solid less than it, viz. to the solid Q_: And divide the pyramid DEFH into two equal 'pyramids, similar to the whole, and into two equal •:.12. prisms. Therefore these two prisms are greater^ than the half of the whol? pyramid. And again, let the pyramids made by this division be in like manner divided, and so on, until the pyramids which -remain undivided in the pyramid DEFH be, all of them together, less than the excess of the pyramid DEFH above the solid Q_: Let these, for example, be the pyramids DPRS, STYH : Therefore the prisms, which make the rest of the pyramid DEFH, are greater than the solid Q^: Divide, likewise the pyramid ABCG in the same manner, and into as many parts, as the pyramid DEFH : Therefore as the base • *•• ABC to the base DEF, so'' are the prisms in the pyramid ABCG to the prisms in the pyramid DEFH : But as the base ABC tothe base DEF, so, by hypothesis, is the pyramid ABCG to the solid Q_; and therefore, as the pyramid ABCG to the solid Q_, so are the prisms in the pyramid ABCG to the prisms in the pyramid DKFH: But the pyramid ABCG is greater " ^^- ^- than the prisms contained in it ; wherefore*^ also the solid Qjs greater than the prisms in the pyramid DEFH. But it is also less, which is impossible. Therefore the bas6 ABC is not to the ♦ This may be explained ll-.e same way as at the note f iu Proposition 2. in th^ Kke cajc. OF EUCLID. 255 the base DEF, as the pyramid ABCG to any solid which is BooxXil. less than the pyramid DEFH. In the same manner it may v^v*^' be demonstrated, that the base DEF is not to the base ABC, as the pyramid DEFH to any solid which Is less than the pyra- mid ABCG. Nor can the base ABC be to the base DEF, as the pyramid ABCG to any solid which is greater than the pyramid DEFH. For if it be possible, let it be so to a greater, viz. thesolid Z. And because the base ABC ist»the base DBF as the pyramid ABCG to thesolidZ ; by inversion, as the base DEF to the base ABC, so is the solid Z to the pyramid ABCG. But as thesolid Z is to the pyramid ABCG, so is tfte pyramid K \ a \ \ \ • \ 2 N \ DEFH to some solid*, which mu>.t be Icss^ than the pyramid * ^^• ABCG, because the solid Z is greater than the pyramid DEFH, And therefore, as the base DEF to the base ABC, io is the pyramid DEFH to a solid less than the pyramid ABCG ; the contrary to which has been proved. Therefore the base ABC is not to the base DEFH, as the pyramid ABCG to any solid which is greater than the pyramid DEFH. And it has been proved, that neither is the base ABC to the base DEF, as the pyramid ABCG to any solid which is less than the pyramid DEFH. Therefore, as the base ABC is to the kase DEF, so is the pyramid ABCG to the pyramid DEFH, Wherefore pyramids, &c. Q_ E, D. * This asay be upbioed the sane vgy as the Tike at fU iMrk f i» frf. 2. 256 Book XII. THE ELEMENTS SeeN. •5. 12. PROP. VI. THEOR. i Y RAM I Ds of the same altitude which have poly- gons for their bases, are to one another as their bases. Letthepyramids which havethepolygonsABCDE,FGHKL for their bases, and their vertices in the points M, N be of the same altitude : As the base ABCDE to the base FGHKL, so is the pyramid ABCDEiM to the pyramid FGHKLN. Divide the base ABCDE into the triangles ABC, ACD, ADE ; and the base FGHKL into the triangles fGH,FHK, FKL : And upon the bases ABC, ACD, ADE let there be as many pyramids of which the common vertex is the point M, and upon the remaining bases as many pyramids having their common vertex in the point N: Therefore since the triangle ABC is to the triangle FGH, as* the pyramid ABCM to the pyramid FGHN ; and the triangle ACD to thetriangle FGH*, as the pyramid ACDM to the pyramid FGHN ; and also the »2 Co D F triangle ADE to the triangle FGH, as the pyramid ADEMto the pyramid FGHN ; as all the first antecedents to their com- mon consequent ; so'' are all the other antecedents to their com- mon consequent ; that is, as the base AfiCDE to the base FGH, so is the pyramid ABCDEM to the pyramid FGHN ; And for the same reason, as the base FGHKL tothe base FGH, so is the pyramid FGHKLN tothe pyramid FGHN : And, by inversion, as the base FGH to the base FGHKL, so is the py- ramid FGHN tothepyramidFGHKLN; Then, because as the base ABCDE to the base FGH, so is the pyramid ABCDEM to the pyrarhid FGHN ; and as the base FGH to the base FGHKL,sois the pyramid FGHN to the pyramid FGHKLN; therefore OF EUCLID. 257 therefore, ex jequali% as the base ABCDE to the base '^ookXII. FGHKL, so the pyramid ABCDEM to the pyramid ^"^^^ FGHKLN. Therefore pyramids, &c. Q^E. D. PROP. VII. THEOR. xirfVERY prism having a triangular base may be divided into three pyramids that have triangular bases, and are equal to one another. Let there be a prism of which the base is the triangle ABC, and let DEF be the triangle opposite to it : Tlie prism ABCDEF may be divided into three equal pyramids having triangular bases. Join BD, EC, CD ; and because ABED is a parallelogram of which BD is the diameter, the triangle ABD is equal* to »si. ]. the triangle EBD ; therefore the pyramid of which the base is the triangle ABD, and veitex the point C, is equal*" to the ^ 5. 12. pyramid of which the base is the triangle EBD, and vertex the point C : But this pyramid is the same with the pyramid the base of which is the triangle EBC, and vertex the point D ; for they are contained by the same planes : Therefore the pyramid of which the base is the triangle ABD, and vertex the point C, is equal to the pyramid, the base of which is the tri- angle EBC, and vertex the point D : Again, because FCBE is a parallelogram of which the diameter is CE, the triangle ECF is equal* to the triangle FCB j therefore the pyramid of which the base is the triangle ECB, and vertex the point D, is equal to the pyra- mid the base of which is the triangle ECF, and vertex the point D : But the pyramid of which the base is the triangle ECB, and vertex the point D, has been proved equal to the pyramid of which the base is the triansrle ABD, and vertex the point C. Therefore the prism ABCDEF is divided into three equal pyramids having tri- angular bases, viz. into the pyramids ABDC, EBDC, ECED: And because the pyramid of which the base is the triangle ABD, and vertex the point C, is the same with the pyramid of which the base is the triangle ABC, and vertex the point D, for they are contained by the same planesj and that the pyramid of which the base is the triangle ABD, and vertex the point C, has been S demonstrated ~ 25^ THE ELEMENTS Book XII. demonstrated to be a third part of the prism, the base of which ^'^"^^'^'^ is the triangle ABC, and to which DEF is the opposite triangle ; therefore the pyramid of which the base is the tri- angle ABC, and vertex the point D, is the third part of the prism which has the same base, viz. the triangle ABC, and DEF is the opposite triangle. Q. E. D. Cor. I. From this it is manifest, that every pyramid is the third part of a prism which has the same base, and is of an equal altitude with it; for if the base of the prism be any other figure than a triangle, it may be divided into prisms having triangular bases. Cor. 2. Prisms of equal altitudes are to one another as their bases ; because the pyramids upon the same bases, and of = 6. 12. the same altitude, are«= to one another as their bases. PROP. VIII. THEOR. oIMILAR pyramids, having triangular bases, are one to another in the triplicate ratio of that of their homologous sides. Let the pyramids having the triangles ABC, DEF for their bases, and the points G, H for their vertices, be similar, and similarly situated ; the pyramid ABCG has to the pyramid DEFH, the triplicate ratio of that which the side BC has to the homologous side EF. Complete the parallelograms ABCM, GBCN, ABGK, and the solid parallelopiped BGML contained by these planes and X those opposite to them: And, in like manner, complete the solid parallelopiped EHPO contained by the three parallelograms DEFP, HEFR, DEHX, and those opposite to them : And because OF EUCLID. 259 because the pyramid ABCG is similar to the pyramid DEFH, ^°»^ xii. the angle ABC is equal * to the angle DEF, and the angle .YTd^Tii GBC to the angle HEF and ABG to DEH : And AB is'' to";. d«f.<. BC, as DE to EF; that is, the sides about the equal angles are proportionals ; wherefore the parallelogram BM is similar toEP : For the same reason, the parallelogram BN is similar to ER, and BK to EX : Therefore the three parallelograms BM,BN, BKl are similar to the three EP, ER, EX : But the ' three BM, BN, BK, are equal and similar*^ to the three which c 24. 11. are opposite to them, and the three EP, ER, EX equal and similar to the three opposite to them : Wherefore the solids BGML, EHPO are contained by the same number of similar planes ; and their solid angles are equal"* ; and therefore thed b ^ solid BGML is similarHothe solid EHPO : But similar solid parallelopipeds have the triplicate •" ratio of that which theiresg, ]]_ homologous sides have : Therefore the solid BGML has to the solid EHPO the triplicate ratio of that which the side BC has to the homologous side EF : But as the solid BGML is to the solid EHPO, so is '^ the pyramid ABCG to the pyramid f 15. 5. DEFH ; because the pyramids are the sixth part of the solids, since the prism, which is the half§ of the solid parallelopiped, e2S. ii. is triple ^ of the pyramid. Wherefore likewise the pyramid » 7. 12. ABCG has to the pyramid DEFH, the triplicate ratio of that which BC has to the homologous side EF, Q. E. D. Cor. From this it is evident, that similar pyramids which Sec J«'. have multangular bases, are likewise to one another in the triplicate ratio of their homologous sides : For they may be divided into similar pyramids having triangular bases, because the similar polygons, which are their bases, may be divided into the same number of similar triangles homologous to the whole polygons ; therefore as one of the triangular pyramids in the first multangular pyramid is to one of the triangular pyramids in the other, so are all the triangular pyramids in the first to all the triangular pyramids in the other j that is, so is the first multangular pyramid to the other : But one triangular pyramid is to its similar triangular pyramid, in the triplicate ratio of their homologous sides ; and therefore the first mul- tangular pyramid has to the other, the triplicate ratio of that which one of the sides of the first has to the homologous •ides of the other. S2 :6o THE elem:e:nts ;.X XII- PROP. IX. THEOR. 1 HE bases aiicl altitudes of equal pyramids.liav- ing triangular bases are reciprocally proportional : And triangular pyrajiiids of which the bases and altitudes are reciprocally proportional, are equal to one another. Let the pyramids of which the triangles ABC, DEF, are the bases, and which have their vertices in the points G, H, be equal to one another ; The bases and altitudes of the pyramids ABCG, DEFH are reciprocally proportional, viz. the base ABC is to the base DEF, as the altitude of the pyramid .DEFFI to the altitude of the pyramid ABCG. Complete the parallelograms AC, AG, GC, DF, DH, HF ; and- the solid^ parallelopipeds BGML, FHPO, contained by •tliese planes and those opposite to them : And because the pyramid ABCG is eq«al to the pyramid DEFH, and that the solid BGML is sextuple of the pyramid ABCG, and the solid EHPO sextuple of the pyramid DEFH ; therefore the solid ' 1. Ax.5. BGML is equal * to the solid EHPO : Put the bases and alti- tudes of equal solid parallelopipeds are reciprocally propor- *3\. \\. tionalb; therefore as the base BM to the base FP, so is the- al- titude of the solid EHPO to the altitude of the solid BGML : But as the base BM to the base EP, so is"^ the triangle ABC to the triangle DEF ; therefore as the triangle ABC to the tri- angle DEF, so is the altitude of the solid EHPO to the alti- tude of the solid BGML: But the altitude of the solid EHPO is the same with the altitude of the pyramid DEFH ; and the altitude of the s»lid BGML is the same with the altitude of the pyramid ' 15. 5. OF EUCLID. 261 pyramid ABCG: Therefore, as the base ABC to the base DEF, 8ooK^?tII. so is' the altitude of the pyramid DEFH to the altitude of the ''*''^''**^ pyramid ABCG : Wherefore the bases and altitudes of the pyramids ABCG, DEFH are reciprocally proportional. Again, let the bases and altitudes of the pyramids ABCG, DEFH be reciprocally proportional, viz. the base ABC to the base DEF, as the altitude of the pyramid DEFH to the alti- tude of the pyramid ABCG : The pyramid ABCG is equal to the pyramid DEFH. The same construction bejng made, because as the base ABC to th J base DEF, so is the altitude of the pyramid DEFH to the altitude of the pyramid ABCG : And as the base ABC to the base DEF, so is the parallelogram BM to the parallelo- gram EP ; therefore the parallelogram BM is to EP, as the altitude of the pyramid DEFH to the altitude of the pyramid ABCG : But the altitude of the pyramid DEFH is the same with the altitude of the solid parallelopiped EHPOi and the altitude of the pyramid ABCG is the same with the altitude of the solid parallelopiped BGML : As, therefore, the base BM to the base EP, so is the altitude of the solid parallelopiped EHPO to the altitude of the solid parallelopiped BGML. But solid parallelopipeds having; their bases and altitudes reci- procally proportional, are equal*" to one another. Therefore " 34. 11. the solid parallelopiped BGML is equal to the solid parallelo- piped EHPO. And the pyramid ABCG is the sixth part of the solid BGML, and the pyramid DEFfI is the sixth part of the solid EHPO. Therefore the pyramid ABCG is equal to the pyramid DEFH. Therefore the bases, &c. Q. E. D. PROP. X. THEOR. JiiVERY cone is the third part of acyhnder which has the same hase, and is of an equal altitude with it. Let a cone have the same base with a cylinder, viz. the cir- cle ABCD, and the same altitude. The cone is the third part of the cylinder ; that is, the cylinder is triple of the cone. If the cylinder be not triple of the cone, it must either be ■' greater than the triple, or less than it. First, Let it be greater "^f than the triple; and describe the square ABCD in the circle: this square is greater than the half of the circl« ABCD*. S 3 Upoiv * As \rai «hewii in Pro^. 2 gf tlu» B«ok. 262 THE ELEMENTS Book XII. Upon the Square ABCDere£l a prism of the same altitiade with ^^^"^ the cylinder ; this prism is greater than half of the cylinder j because if a square be described about the circle, and a prism ereiSled upon the square, of the same altitude with the cylinder, the inscribed square is half of that circumscribed ; and upon these square bases are ere6led solid parallelopipeds, viz. the prisms of the same altitude ; therefore ' the prism upoh the square ABCD is the half of the prism upon the square described about the circle : Because they are to one another as their > 3* 1 1, bases* : And the cylinder is less than the prism upon the square described about the circle ABCD: Therefore the prism upon the square ABCD of the same altitude with the cylinder, is greater than half of the cylinder. Bisedt the circumferences AB, BC, CD, DA in the points E, F, G, H ; and join AE, EB, BF, FC, CG, GD, DH, HA : Then, each of the triangles AEB, BFC, CGD, DHA is greater than the half of the seg- ment of the circle in which it stands, as was shewn in Prop. 2. of this Book. Ere£t prisms upon each of these triangles of the same altitude with the cylinder; each- of these prisms is greater than half of the seg- ment of the cylinder in which it is ; because if, through the points E, F, G, H, parallels be drawn to AB, BC, CD, DA, and parallelograms be completed upon the same AB, BC, CD, DA, and solid parallelopipeds be eredled upon the parallelograms ; the prisms upon the tri- angles AEB, BFC, CGD, DHA are the halves of the solid ► '2 Cor. parallelopipeds''. And the segments o( the cylinder which arc 7. u. upon the segments of the circle cut ofFby AB, BC, CD, DA, are less than the solid parallelopipeds which contai.n them. Therefore the prism3 upon the triangles AEB, BFC, CGD, DHA, are greater than half of the segments of the cylinder in which they are j therefore, if each of the circumferences be divided into two equal parts, and straight lines be drawn from the points of division to the extremities of the circumferences, and upon the triangles thus made, prisms be eredled of the same altitude with the cylinder, and so on, there must at length Lcmms. remain some segments of the cylinder which together are less,'^ than the excess of the cylinder" above the triple of the cone, Let them be those upon the segments of the circle AE, EB, BF. FC, OF EUCLID. 263 19. FC, CG, GD, DH, HA. Therefore the rest of the cylin- BoorXII. der, that is, the prism of which the base is the polygon AEBFCGDH, and of which the altitude is the same with that of the cylinder, is greater than the triple of the cone : But this prism is triple"* of the pyramid upon the same base, of which * 1 Cor. T. the vertex is the same with the vertex of the cone; therefore the pyramid upon the base AEBFCGDH, having the same vertex with the cone, is greater than the cone, of which the base is the circle ABCD : But it is also less, for the pyramid is contained within the cone ; which is impossible. Nor can the cylinder be less than the triple of the cone. Let it be less, if possible ; therefore, inversely, the cone is <Treater than the third part of the cylinder. In the circle ABCD describe a square ; this square is greater than the half of the circle; And upon the square ABCD ereft a p) ramid, having the same vertex with the cone: this pyramid is greater than the halfof the cone j because, as was before demonstrated, if a square be described about the circle, the square ABCD is the half of it; and if upon these squares there be erected solid parallelo- pipeds of thesame altitude wih the cone, which are also prisms, the prism upon the square ABCD shall be the half of that which is upon the square, described about the circle; for they are to one another as their bases'^ ; as are also the third parts of them : Therefore the pyramid, the base of which is the square ABCD, is half of the pyramid upon the square de- scribed about the circle : But this last pyramid is greater than the cone which it contains ; therefore the pyramid upon the square ABCD, having the same vertex with the cone, is greater than the half of the cone. Bisect the circumferences AB, BC ; CD, DA in the points E, F, G, H, and join AE, EB, BF, FC, CG, GD, DH, HA: Therefore each of the triangles AEB, BFC, CGD, DHA is greater than half of the segment of the circle in which it is ; Upon each of these triangles ereft pyramids having the same vertex with the cone. Therefore each of those pyramids is greater than the half of the segment of the cone in which it is, as before was demonstrated of the prisms and segments of the cylinder ; and thus dividing each ©f the circumferences into two equal parts, and joining the S 4 points « 3'i. 11. 264 THE ELEMENTS liooK XII. points of division and their extremities by straight lines, and "'^"''''^'^ upon the triangles erecting pyramids having their vertices the same with that of the cone, and so on, there must at length remain some segments of the cone, which together shall be less than the excess of the cone, above the third part of the cylinder. Let these be the segments upon AE, EB, BF, FC, CG, GD, DH, HA. Therefore the rest of the cone, that is, the pyramid, of which the base is the polygon AEBFCGDH, and of which the vertex is the same with that of the cone, is greater than the third part of the cylinder. But this pyramid is the third part of the prism upon the same base AEBFCGDH, and of the same altitude with the cylin- der. Therefore this prism is great- er than the cylinder of which the base is the circle A BCD. But it is also less, for it is contained within the cylinder ; which is impossible. Therefore the c'y- , linder is not less than the triple of the cone. And it has been demonstrated that neither is it greater than the triple. There- fore the cylinder is triple of the cone, or, the cone is the third part of the cylinder. Wherefore every cone, &c. C^ E. D. Sf« N. C PROP. XL THEOR. ONES and cylinders of the same altitude, are to one another as their bases. Let the cones and cylinders, of which the bases are the cir- cles ABCD,EFGH, and the axes KL, MN, and AC, EG the diameters of their bases, be of the same altitude. As the circle A BC D to the circle EFGH, so is the cone AL to the cone EN. If it be not so, let the circle ABCD be to the circle EFGH, as the cone AL to some solid either less than the cone EN, or greater than it. First, let it be to a solid less than EN, viz. to the solid X ; and let Z be the solid which is equal to the excess of the cone EN above the solid X ; therefore the cone EN is equal to the solids X, Z together. In the circle EFGH describe the square EFGH, therefore this square is greater than the half of the circle : Upon the square EFGH eredl a pyra- mid of the same altitude with the cone ; this pyramid is greaiter than half of the cone. For, if a square be de- scribed about the circle, and a pyramid be ere^ed upon it, having OF EUCLID. 265 having the same vertex with the cone*, the pyramid inscribed book xii. in the cone is half to the pyramid circumscribed about it, *^—n^^^ because they are to one another as their bases* : But the cone »6. 12. is less than the circumscribed pyramid ; therefore the pyramid of which the base is the square EFGH, and its vertex the same with that of the cone, is greater than half of the cone: Divide the circumferences EF, FG, GH, HE, each into two equal parts in the points O, P, R, S, and join EO, OF, FP, PG, GR, RH, HS, SE : Therefore each of the triangles EOF, FPG, GRH, HSE is greater than half of the segment of the \ N \ \ I X ^ \ n \. V N \i z \ circle in which it is : Upon each of these triangles ere£l; a py- ramid having the same vertex with the cone ; each of these pyramids is greater than the half of the segment of the cone in which it is : And thus dividing each of these circumferences into two equal parts, and from the points of division drawing straight lines to the extremities of the circumference, and up- on each of the triangles thus made erecting pyramids, having the same vertex with the cone, and so on, there must at length remain some segments of the cone which are together less** * Lea. than the solid Z: Let these be the segments upon EO, OF, FP, PG, • Vertex is put in place of aUitude, which is in the Qreek, because the pjTjmid, ia ■what follows, it supposed to be circumscribed abcjt the cone, and so must have thK »»inc fcrtex. Aad the ssae cbang* it waas in scne place: ibUo'Vf ing. 1 266 THE ELEMENTS Book XII. pQ, QR, RH, HS,SE: Therefore the remainder of the cone, ^"^ viz. thepyramidofwhichthebaseisthepolygon EOFPGRHS, and its vertex the same with that of the cone, is greater than the solid X : In the circle ABCD describe the polygon ATBYC VDQ^similartothe polygon EOFPGRHS,and upon it erect a pyramid having the same vertex w^ith the cone AL : And because as the square of AC is to the square of EG, so^ is the polygon ATBYCVDQ^to the polygon EOFPGRHS; and as the square of AC to the square of EG, so is*" the circle ABCD to the circle EFGH ; therefore the circle ABCD*^ is to the circle EFGH,asseepolygonATBYCVDQ to the polygon »1. 12. " 2. 12. <= 11.5. C E I f\ [\ X \ \ N \ \ \ z EOFPGRHS : But as the circle ABCD to the circle EFGH, so is the cone AL to the solid X: and as the polygon *^' ATB YC VDQ^to the polygon EOFPGRHS, sd is" the pyra- mid of vi^hich the base is the first of these polygons, and vertex L, to the pyramid of which the base is the other polygon, and its vertex N : Therefore, as the cone AL to the solid X, so is the pyramid of which the base is the polygon ATBYC VDQ^, and vertex L, to the pyramid the base of which is the polygon j4 3 EOFPGRHS, ajid vertex N : But the cone AL is greater than the pyramid contained in it ; therefore the solid X is greater'^ than the pyramid in the cone EN. But it isless, as was shewn, whicft OF EUCLID. 267 which is absurd : Therefore the circle ABCD is not to the Book xir. circle EFGH, as the cone AL to any solid which is less than ^■^^•^*-' the cone EN. In the same manner it may be demonstrated, that the circle EFGH is not to the circle ABCD, as the cone EN to any solid less than the cone AL. Nor can the circle ABCD be to the circle EFGH, as the cone AL to any solid greater than the cone EN : For, if it be possible, let it be so to the solid I, which is greater than the cone EN : Therefore, by inversion, as the circle EFGH to the circle ABCD, so is the solid I to the cone AL : But as the solid I to the cone AL, so is the cone EN to some solid, which must be less» than the cone AL, because the solid I is greater than » u. 3t the cone EN : Therefore, as the circle EFGH is to the circle ABCD, so is the cone EN to a solid less than the cone AL, which was shewn to be impossible : Therefore the circle ABCD is not to the circle EFGH, as the cone AL is to any solid greater than the cone EN : And it has been demonstra- ted, that neither is the circle A^CD to the circle EFGH, as the cone AL to any solid less than the cone EN : Therefore the circle ABCD is to the circle EFGH, as the cone AL to the cone EN : But as the cone is to the cone, so'' is the cy- jj 5 linder to the cylinder, because the cylinders are triple'^ of the cone, each to each. Therefore as the circle ABCD to the circle EFGH, so are the cylinders upon them of the same altitude. Wherefore cones and cylinders of the same altitude are to one another a,s their bases. Q. E. D. PROP. XII. THEOR. Similar cones and cylinders have to one ano- seeN. ther the triplicate ratio of that which the diameters of their bases have. Let the cones and cylinders of which the bases are the cir- cles ABCD, EFGH, and the diameters of the bases AC, EG, and KL, MN, the axis of the cones or cylinders, be similar : The cone of which the base is the circle ABCD, and vertex the point L, has to the one of which the base is the circle EFGH, and vertex N, the triplicate ratio of that which AC has to EG. For if the cone ABCDL has not to the cone EFGHN the triplicate ratio of that which AC has to EG, the cone ABCDL shall have the triplicate of that ratio to some solid Which is less or greater than the cone EFGHN. First, let it have it to a less, viz. = 10. 13. 268 THE. ELEMENTS Book xn. viz to the solid X. Make the same construction as in the pre- '"^^ ceding proposition, and it may be d-monstrated the very same way as in that proposition, that the pyramid of which the base is the polygon EOFPGRHS, and vertex N, is greater than the solid X. Describe also in the circle ABCD the polygon ATBYCVDQ similar to the polygon EOFPGRHS, upon which ere6l a pyramid having the same vertex with the cone ; and let LAQ^ be one of the triangles containing the pyra- mid upon the polygon ATBYCVDQ^, the vertex of which is L ; and let NEb be one of the triangles containing the pyramid upon the polygon EOFPGRHS of which the ver- tex is N J and join KQ^, MS : Because then the- cone ='24def.ii. ABCDL is similar to the cone EFGHN, AC is' to EG as ■i>i5. 5. the axis KL to the axis MN ; and as AC to EG, so* is AK to EM ; therefore as AK to EM, so is KL to MN ; and, alternately, AK to KL, as EM to MN : And the right angles AKL, EMN are equal j therefore the sides about these equal angles beirig proportionals, the triangle AKL is similar*^ to the triangle EMN. Again, because AK is to KQ,, as EM to MS, and that these sides are about « (kG. OF EUCLID. 269 about equal . angles AKQj EMS, because tbese angles are, BookXII, each of them, the same part of four right angles at the centres K, IVl ; therefore the triangle AKQ_is similar^ to the triangle a g. g. EiVlS : AnJ because it has been shown that as AK to KI^ so is EM to MN, and that AK is equal to KQ_; and EM to MS, as QK to KL, so is SM to MN : and therefore the sides about the right angles QKL, SMN being proportionals, the triangle LKQ^is similar to the triangle NMS : and because of the similarity of the triangles AKL, EMN, as LA is to AK, so is NE to EM; and by the similarity of the triangles AKQ_, EMS, as KA to AQ^, so ME to ES ; ex lequali^ LA is " --• ^ to AQ, as NE to ES. Again, because of the similarity of the triangles LQK, NSM, as LQ, to QK, so NS to SM ; and from the similarity of the triangles KAQ_, MES, as KQ^to QA, so MS to SE ; ex aequali'', LQ is to Q_\, as NS to SE : And it was proved that QA is to AL, as SE to EN ; there- fore, again, ex sequali as QL to LA, so is SN to NE : Where- fore the triangles LQ_A, NSE, having the sides about all their angles proportionals, are equiangular' and similar to one'^S-fi- another : And therefore the pyramid of which the base is the triangle AKQ_, and vertex L, is similar to the pyramid the base of which is the triangle EMS, and vertex N, because their solid angles are equal** to one another, and they are contained ''B. 11. by the same number of similar planes: But similar pyramids which have- triangular bases have to one another the triplicate 'ratio of that which their homologous sides have; therefore *s. 12. the pyramid AKQL has to the pyramid EMSN the triplicate ratio of that which AK has to EM. In the same manner, if straight lines be drawn from the points D, V, C,"Y, B, T, to K, and from the points H, R, G, P, F, O, to M, and py- ramids be erected upon the triangles having the same vertices with the cones, it may be demonstrated that each pyramid in the first cone has to each in the other, taking them in the same order, the triplicate ratio of that which the side AK hastothc side EM ; that is, which AC has to EG : But as one antece- dent to its consequent, so are all the antecedents to all the consequents*^ ; therefore as the pyramid AKQL to the pyra- ^ 12. 5. mid EMSN, so is the whole pyramid the base of which is the polygon DQATBYCV, and vertex L, to the whole pyramid of which the base is the polygon HSEOFPGR,and vertex N. Wherefore also the first of these two last-named pyramids has to the other the triplicate ratio of that which AC has to EG, But, by the hypothesis, the cone of which the base is the cir- cle ABCD, and vertex L, has to the solid X, the triplicate ratio of that which AC has to EG ; therefore, as ihe cone of which 270 THE ELEMENTS Book xri. which the base is the circle ABCD, and vertex L, is to the ^"""'''^'^ solid X, so is the pyramid the base of which is the polygon DQA i'-BYCV, and vertex L, to the pyramid the base of which is the polygon HSEOFPGR and vertex N : But the said cone ^ is greater than the pyramid contained in it, therefore the solid X is greater^ than the pyramid, the base of which is the poly- gon HSEOFPGR, and vertex N ; but it is also less, which is impossible : therefore the cone, of which the base is the circle u \ \ X \ ABCD and vertex L, has not to any solid which is less than the cone of which the base is thecircle EFGH and vertexN,the tri- plicate ratio of that which AC has to EG. In the same manner it may be demonstrated, that neither has the cone EFGHN to any solid which is less than the cone ABCDL, the triplicate ratio of that which EG has to AC. Nor can the cone A BCDL have to any solid which is greater than the cone EFGHN, the j triplicate ratio of that which AC has to EG : For, if it be pos- sible, let it have it to a greater, viz. to the solid Z : Therefore, inversely, the solid Z has to the cone ABCDL, the triplicate ratio of that which EG has to AC : But as the solid Z is to the OF EUCLID. 271 thecone ABCDL, soistheconeEFGHN to some solid," which BookXil tnust be less* than the cone ABCDL, because the solid Z is , ^^ ^ greater than the cone EFGHN : Therefore the cone EFGHN has to a solid which is less than the cone ABCDL, the tripli- cate ratio of that which EG has to AC, which was demon- strated to be impossible : therefore the cone ABCDL has not to any solid greater than the cone EFGHN, the triplicate ra- tio of that which AC has to EG ; and it was demonstrated, that it could not have that ratio to any solid less thaa the cone EFGHN : Therefore the cone ABCDL has to the cone EFGHN, the triplicate ratio of that which AC has to EG : But as the cone is to the cone, so'' the cylinder to the cylinder; ' for every cone is the third part of the cylinder upon the same base, and of the same altitude ; Therefore also the cylinder has to the cylinder, the triplicate ratio of that v/hich AC has to EG: Wherefore similar cones, Sec. Q. E. D. 15. 5. PROP. Xin. THEOR. -IF a cylinder be cut by a plane parallel to its oppo- See n. site planes, or bases ; it divides the cylinder into two cylinders, one of which is to the other as the axis of the first to the axis of the other. Let the cylinder AD be cut by the plane GH parallel to the opposite planes AB, CD, meeting the axis EF in the point K, and let the line GH be the common section of the plane GHand the surface of the cylin- der AD : Let AEFC be the paralle- logram in any position of it, by the revolution of which about the straight line EF the cylinder AD is described : and let GK be the common section of the plane GH, and the plane AEFC: And because the parallel planes AB, GH, are cut by the plane AEKG, AE, KG, their commoa sections with it are parallel* ; where- fore AK is a parallelogram, and GK equal to FA the straight line from the centre of the circle AB : For the same reason, each of the straight lines K A S B H D Y drawn 16, II, 272 EooK'XlI, •15. def.l * 11.12. THE ELEMENTS drawn from the point K to the line GH may be proved to be equal to those which are drawn from the centre of the circle AB to its circumference, and are therefore all equal to one another. Therefore the line GH is the circumference of a circle* j of which the centre is the point K : Therefore the plane GH divides the cylinder AD into the cylinders AH, GD ; for they are the same which would be described by the revolution of the parallelograms AK, GF, about the straight lines EK, KF: And it is to be shewn, that the cylinder AH is to the cylin- der HC, as the axis EK to the axis KF. Produce the axis EF both ways ; and take any number of Straight lines EN, NL, each equal to EK } and any number EX, XM, each equal to tK ; and let planes parallel toAB, CD pass through the points L, N, X, M : Therefore the common sections of these planes with the cylinder produced are circles the centres of which are the points L, N, X, M, as was proved of the plane GH ; and these planes cut off the cylinders PR, RB, DT, TQ : And because the axes LN, NE, EK are all equal; therefore the cylinders PR, RB, BG are'' to one another as their bases ; but their bases are equal, and therefore the cylinders PR, RB, BG are equal : And because the axes LN, NE, EK are equal to one ano- ther, as also the cylinders PR, RB, BG,, and that there are as many axes as cylinders ; therefore, whatever multiple the axis KL is of the axis KE, the same multiple is the cylinder PG of the cy- linder GB: ¥oT the same reason, whatever multiple the axis MK is of the axis KF, the same multiple is the cylinder QG of the cylinder GD : And if the axis KL be equal to the axis KM, the cylinder PG is equal to the cylinder GQ ; and if the axis KL be greater than the axis KM, the cylinder PG is greater than the cylinder QG ; and if less, less : Since, there- fore there are four magnitudes, viz. the axis EK, KF, and the cylinders BG, GD, and that of the axis EK and cylinder JJG, there has been taken any equimultiples whatever,viz. the axis OF EUCLID. 273 axis KL and cylinder PG ; and of the axis KF and cylinder Book xil. GD, any equimultiples whatever, viz. the axis KM an<? ^^^^^"""^ ■ cylinder GQ; arid it has been demonstrated, if the axis KLbe greater than the axis KM, the cylinder PG is greater than the cylinder GQ ; and if equal, equal ; and if less, less: Therefore •"the axis EK is to the axis KF, as the cylinder BG to the«5. def. r cylinder GD. Wherefore, if a cylinder, &c. Q. E. D. PROP. XIV. THEOR. v^ONES and cylinders upon equal bases are to one another as their altitudes. Let the cylinders EB, FD be upon the equal bases AB, CD : As the cylinder EB to the cylinder FD, so is the axis GH to the axis KL, Produce the axis KL to the point N, and make LN equal to the axis GH, and let CM be a cylinder of which the base is CD, and axis LN j and because the cylinders EB, CM have the same altitude, they are to one another as their bases' : But * n. ]o^ their bases are equal, therefore also the cylinders EB, CM are equah And because the cylin- der FM is cut by the plane CD parallel to its opposite planes, as the cylinder CM to the cylinder FD so is* the axis LN to the axis KL. But the cylinder CM is equal to the cylinder EB, and the axis LN to the axis GH : There- fore as the cylinder 'EB to the cylinder FD, so is the axis GH to the axis KL : And as the cylinder EB to the cylinder FD, so is«= the cone ABG to c 15. 5^ the cone CDK, because the cylinders are triple' of the cones : 4 10. ig. Therefore also the axis GH is to the axis KL, as the cone ABG to the cone CDK, and the cylinder EB to the cylinder FD. Wherefore cones, &c, Q^ E. D. k 13. 12.- 274 Book XIl. THE ELEMENTS ie«N. PROP. XV. THEOR. 1 HE bases and altitudes of equal cones and cylin- ders, are reciprocally proportional ; and if the bases and altitudes be reciprocally proportional, the cones and cylinders are equal to one another. Let the circles A BCD, EFGH, the diameters of which are AC, EG, be the bases, and KL, MN the axis, as also the al- titudes, of equal cones and cylinders ; and let ALC, ENG be the cones, and AX, EO the cylinders : The bases and alti- tudes of the cylinders AX, EO are reciprocally proportional ; that is, as the base ABCD to the base EFGfi, so is the alti- tude MN to the altitude KL. Either the altitude MN is equal to the altitude KL, or these altitudes are not equal. First, let them be equal ; and the cylin- ders AX, EO being also equal, and cones and cylinders of the -»n. 12. same altitude being to one another as their bases*, therefore b A. 5. the base ABCD is equal'' to the base EFGH ; and as the base ABCD is to the base EFGH, so is the altitude MN to the altitude KL. But let the alti- tudes KL, MN, be unequal, and MN the greater of the two, and from M'N take MP equal to KL, and through the point P cut the cylinder EO by the plane TYS, parallel to the opposite planes of the circles EFGH, RO ; therefore the common sedtion of the plane TYS and the cylinder EO is a circle, and consequently ES is a cylinder, the base of which is the circle EFGH, and altitude MP : And because the cy- linder AX is equal to the cylinder EO, as AX is to the cylin- ' '• ^' der ES, so « is the cylinder EO to the same ES : But as the cylinder AX to the cylinder ES, so* is the base ABCD to the base EFGH ; for the cylinders AX, ES are of the same * 13. 12. altitude ; and as the cylinder EO to the cylinder ES, so* is the altitude MN to the altitude MP, because the cylinder EO OF EUCLID. 275 EO is cut by the plane TYS parallel to its opposite planes. 2=°^ ^^^• Therefore as the base ABCD to the base EFGH, so is the '*'*^***^ altitude MN to the altitude MP: But MP is equal to the al- titude KL ; wherefore as the base ABCD to the base EFGH, so is the altitude iMN to the altitude KL ; that is, the bases and altitudes of the equal cylinders AX, EO are reciprocally proportional. But let the bases and altitudes of the cylinders AX, EO, be reciprocally proportional, viz. the base ABCD to the base EFGH, as the altitude MN to the altitude KL : The cylin- der AX is equal to the cylinder EO. First, Let the "base ABCD be equal to the base EFGH ; then because as the base ABCD is to the base EFGH, so is the altitude MN to the altitude KL ; MN is equal** to KL, "A. 5. and therefore the cylinder AX is equal' to the cylinder EO. » n..i2. But let the bases ABCD, EFGH be unequal, and let ABCD be the greater ; and because, as ABCD is to the base EFGH, so is the altitude MN to the altitude KL j therefore MN is greater* than KL. Then, the same construction being made as before, because as the base ARCD to the base EFGH, so is the altitude MN to the altitude KL ; and because the altitude KL is equal to the altitude MP ; therefore the base ABCD is* to the base EFGH, as the cylinder AX to the cylinder ES ; and as the altitude MN to the altitude MP or KL, so is the cylinder EO to the cylinder ES : Therefore the cylinder AX is to the cylinder ES, as the cylinder EO is to • the same ES : Whence the cylinder AX is equal to the cylin- der EO i and the same reasoning holds in cones. Q^ E. D. ^ PROP. XVL PROB. 1 O describe in the greater of two circles' that have the same centre, a polygon of an even number of equal sides, that shall not meet the lesser circle. Let ABCD, EFGH be two given cfrcles having the same centre K : It is required to inscribe in the greater circle ABCD, a polygon of an even number of equal sides, that shall not meet the lesser circle. Through the centre K draw the straight iine BD, and from %he point G, where it meets the circuiiiferencei of tfee lesser T 2 circle. 276 THE ELEMENTS "16. 3. Lemn Book XFt. circle, draw GA at right angles to BD, and produce it to C ; therefore AC touches'* the circle EFGH : Then, if the circum- ference BAD be bise6led, and the half of it be again bisedled, and so on, there must at length remain a circumference less^ than AD : Let this be LD ; and from the point L draw LM per- pendicular to BD, and produce it to N i and join LD, DN. Therefore LD is equal to DN j and because LN is parallel to AC, and that AC touches the circle EFGH i therefore LN does not meet the circle EFGH. And much less shall the straight lines LD, DN meet the circle E FGH ; So that if straight lines equal to LD be applied in thecircle ABCD from the point L around to N, there shall be described in the circle a polygon of an even number of equal sides not meeting the lesser circle. Which was to be done. «2S.3, LEMMA IL If two trapeziums ABCD, EFGH be inscribed in the circles, the centres of which are the points K, L ; and if the sides AB, DC be parallel, as also EF, HG ; and the other four sides AD, BC, EH, FG, be all equal to one another ; but the side AB greater than EF, and DC greater than HG. The straight line KA from the centre of the circle in which the greater sides are, is greater than the straight line LE drawn from the centre to the cir- cumference of the other circle. If it be possible, let KA be not greater than LE ; then K A must be either equal to it, or less. First, let KA be equal to LE : Therefore, because in two equal circles AD,BC, in the ©ne, are equal to EH, FG in the other, the circumferences AD, BC are equaU to the circumferences EH, FG; but because the straight lines AB,DC are respedlively greater than EF, GH, the circumferences AB, DC are greater than EF, HG : Therefore the whole circumference ABCD is greater (^an the whole EFGH -, but it is also equal to it, which is inpossible : OF EUCLID. 277 impossible: Therefore the straight line KA is not equal to BookXH. LE. ^ '-"^'^ But let KA be less than LE, and make LM equal toKA, and from the centre L, and distance LM, describe the circle MNOP, meeting the straight lines LE, LP, LG, LH, in M, N, O, P J and join MN, NO, OP, PM, which are respectively parallel '^ to and less than EF, FG, GH, HE : Then > 2. 5. because EH is greater than MP, AD is greater than MP j and the circles ABCD, MNOP are equal ; therefore the circum- ference AD is greater than MP ; for the same reason, tha cir- cumference DC is greater than NO ; and because the straight line AB is greater than EF, which is greater than MN, much more is AB greater than MN : Therefore the circumference AB is greater than MN j and, for the same reason, the cir- cumference DC is greater than PO.: Therefore the whole circumference ABCD is greater than the whole MNOP j but it is likewise equal to it, which is impossible : Therefor-- KA is not less than LE : nor is it equal to it j the straight line KA must therefore be greater than LE. Q^ E. D. Cor. And if there be an isosceles triangle, the sides of which are equal to AD, BC, but its base less than AB the greater of the two sides AB, DC ; the straight line KA may, in the same manner, be demonstrated to be greater than the straight line drawn from the centre to the circumference of the circle described about the triangle. T3 278 THE ELEMENTS PROP. XVII. PROB. seeN. I o describe in the greater of two spheres wliicli have the same centre, a solid polyhedron, the su- perficies of which shall not meet the lesser sphere. Let there be two spheres about the same centre A ; it is required to describe in the greater a solid polyhedron, the superficies of which shall not meet the lesser sphere. Let the spheres be cut by a planepassing through the centre ; the common sections of it with the spheres shall be circles ; because the sphere is described by the revolution of a semicir- cle about the diameter remaining unmoveable ; so that in what- , ever position the semicircle be conceived, the common section of the plane in which it is with the superficies of the sphere is the circumference of a circle : and this is a great circle of the sphere, because the diameter of the sphere, which is likewise » j5 3^ the diameter of the circle, is greater * than any straight line in the circle or sphere: Let then the circle made by the section of the plane with the greater sphere be BCDE, and with the lesser sphere be FGH ; and draw the two diameters BD, CE, at right angles to one another ; and in BCDE, the greater of j^ the two circles, describe '' a polygon of an even number of * ' equalsidesnotmeetingthelesser circle FGH 5 and let its sides, in BE the fourth part of the circle, be BK, KL, LM, ME ; join KAj and produce it to N ; and from A draw AX at right angles to the plane of the circle BCDE, meeting the super- ficies of the sphere in the point X j and let planes pass through AX, and each of the straight lines BD, KN, which, from what has been said, shall producegreat circles on the superficies of the sphere", and let BXD, KXN be the semicircles thus made upon the diameters BD, KN: Therefore, because XA is at right angles to the plane of the circle BCDE, every plane e 18. 11, which passes through X A is at right*^ angles to the plane of the circle BCDE ; wherefore the semicircles BXD, KXN are at right angles to that plane : Anil because the semicircles BED, BXD, KXN, upon the equal diameters BD, KN, are equal to one another, their halves BE, BX, KX, are equal to one another : Therefore, as many sides of the polygon as are in BE, so many there are in BX, KX equal to the sides BK, KL, LM, ME : Let these polygons be described, and their sides be BO, OP, PR, RX s KS, ST, l^Y, YX, and join OS, OF EUCLID. 279 OS, PT, RY ; and from the points O, S draw O V, SQ_per. Book XJI. pendiculars to AB, AK : And because the plane BOXD is at "■^^^^*^ right angles to the plane BCDE, and in one of them BOXD, O V is drawn perpendicular to AB the common seftion of the planes, therefore OV is perpendicular* to the plane BCDE For the same reason J^Q_ is perpendicular to the same plane, because the plane KSXN is at right angles to the plane BCDE. Join VQ; and because in the equal semicircles BXD, KXN »4.def. 11. the circumferences, BO, KS are equal, and OV, SQjare per- pendicular to their diameters, therefore'' OV is equal to SQj ■* 2i>. - and B V equal to KQ^ But the whole B A is equal to the whole KA, therefore the remainder VA is equal to the remainder QA : As therefore B V is to VA, so is KQ^to QA> wherefore VQ^is parallel' to BK: And because OV^, SQ_ are each of '2. 6. them at right angles to the plane of the circle BCDE, OV is parallel^ to SQ_; and it has been proved, that it is also equal '^6. 11 to it ; therefore QV, SO are equal and parallel 8 : And because ' as. 1 Oy is parallel to SO, and also to KB ; OS is parallel'' to BK ; » 9. n and therefore BO, KS which join them are in the same plane . T 4 in iSo THE ELEMENTS SooK. XII, in which these parallel are, and the quadrilateral figure KBOS ^ is in one plane : And if BF, TK be joined, and perpendiculars be drawn from the points P, T to the straight lines AB, AK, it may be demonstras d, that TP is parallel to KB in the very same way that SO was shewn to be parallel to thcsameKBj »>. 11. wher fore* TP is parallel to SO, and the quadrilateral figure SOPT is in one plane; For the same reason the quadrilateral TPRY is in one plane : and the figure YRX is also in one "i. 11. planch Therefore, if from the points O, S, P, T, R, Y, there be drawn straight lines to the point A, there shall be formed a solid polyhedron between the circumferences BX, KX, com- posed of p) ram ids, the bases of which are the quadrilaterals KBOS, SOPT, TPRY, and the triangle YRX,and of which the common vertex is the point A : And if the same ponstruc- tion be made upon each of the sides KL, LM, ME, as has been done upon BK, and the like be done also in the other three quadrants, and in the other hemisphere ; there shall be formed a solid polyhedron described in the sphere, composed «f pyramids, the bases of which arc the aforesaid quadri- lateral OF EUCLID. 2S1 lateral figures, and the triangle YRX, and those formed in BookXu. the lika manner in the rest ot the sphere, the common vertex of them all being the point A: and the superficies of this solid polyhedron does not meet the lesser sphere in which is the circle FGH : For, from the point A draw* AZ perpendicular * n. n, to the plane of the quadrilateral KBOS, meeting it in Z, and join BZ, ZK : And because AZ is perp;:-ndicular to the plane KBOS, it makes right angles with every straight line meeting it in that plane; therefore AZ is perpendicular to BZ and ZK: And because AB is equal to AK, and that tne squares of AZ, ZB, are equal to the square of AB ; and the squares of AZ, ZK to the square of AK''; therefore the sqtiares of AZ, ZB *47. i. are equal to the squares of AZ, ZK : Take from these equals the square of AZ, the remaining square of BZ is equal to the remaining square of ZK ; and therefore the straight line BZ is equal to ZK: In the like manner it may be demonstrated, that the straight lines drawn from the point Z to the points O, S are equal to BZ or ZK : Therefore the circle described from the centre Z. and distance ZB, shall passthrough the points K, O, S, and KBOS shall be a quadrilateral figure in th? circle: And because KB is greater than QV, and QV equal to SO, therefore KB is greater than SO : But KB is equal to each ot thes£raightlinesBO,KS; whereforeeach of the circumferences cut oft by KB, BO, KS is greater than that cut off^by OS; and these three circumferences, together with a fourth equal to one of them, are greater than the same three together with that cut ofl^ by OS ; that is, than the whole circumference of the cir- cle ; therefore the circumference subtended by KB is greater than the fourth part of the whole circumference of the circle KBOS, and consequently the angle BZK at the centre is greater than a right angle: And because the angle BZK is obtuse, thesquare of BK is greater'than the squares of BZ,ZK;' 12. 0. that is, greater than twice the square of BZ. Join KV, and becaus inthe triangles KBV,OBV, KB,BVareequaltoOB, BV, and that they contain equal angles ; the angle KBV is equal"" to the angle OVB : And OVB is a right angle ; there- 4 4, 1. fore also KV'B is a rioht angle : And because BD is less than twice DV; the rectangle contained by DB, BV is less than twice the rectangle Dv B ; that is% the square of KB is less* 8. 6. than twice the square of KV : But the square of KB is greater than twice the square of BZ ; therefore the square of KV is greater than the square of BZ : And because BA is equal to AK, and that the squares ol BZ, Z A are equal together to the square of BA, and the squares of KV, V A to the square of AK i 282 THE ELEMENTS ^ Book xn. ^f^ . therefore the squares of BZ, ZA are equal to the square* ^^'^■'"'^ of K V, VA ; and of tl>ese the square of K V is greater than the square of BZ ; therefore the square of VA is less than the square of ZA, and the straight line AZ greater than VA : Much more then is AZ greater than AG ; because, in the pre- ceding proposition, it was shewn that KV falls without the circle FGH : And AZ is perpendicular to the plane KBOS, and is therefore the shortest of all the straight lines that can be drawn from A, the centre of the sphere to that plane. There- fore the plane KBOS does not meet the lesser sphere. And that the other planes between the quadrants BX, KX fall without the lesser sphere, is thus demonstrated : From the point A draw AI perpendicular to the plane of the quadri- lateral SOPT, and join lO; and, as was demonstrated of the plane KBOS and the point Z, in the same way it may be shewn that the point I is the centre of a circle described about SOPT : and that OS is greater than PT; and PT was shewn to be parallel to OS : Therefore because the two trapeziums KBOS, SOFT inscribed in circles have their sides BK, OS parallel, as also OS, PT; and their other sides BO, KS, OP, ST, all equal to one another, and that BK is greater than OS, and OS • 2.Lem. greater than PT, therefore the straight line ZB is greater' than lO. Join AO which will be equal to AB ; and because AIO, AZB are right angles, the squares of AI, lO are equal to the square of AO or of AB i that is, to the squares of AZ, ZB ; and the square of ZB is greater than the square of lO, therefore the square of AZ is less than the square of AI ; and the straight line AZ less than the straight line AI : And it was proved, that AZ is greater than AG ; much more then is AI greater than AG : Therefore the plane SOPT falls wholly without the lesser sphere : In the same manner it may be de- monstrated, that theplaneTPRY falls withoutthe same sphere, as also the triangle YRX, viz. by the Cor. of 2d Lemma. And after the same way it may be demonstrated, that all the planes which contain the solid polyhedron, fall without the lesser sphere. Therefore ir>the greater of two spheres, which have the same centre, a solid polyhedron is described, the super- fices of which does not meet the lesser sphere. Which was to be done. But the straight line AZ may be demonstrated to be greater than AG otherwise, and in a shorter manner, without the help of Prop. 16, as follows. From the point G draw GU at right angles to AG, and join AU. -If then the circumferences BE bo bise<^d,-and its half again bisefted, and so on, there will at length OF EUCLID. 283 Icngtfibcleftacircumferencelessthaathecircumference which B^jkXII. is subtended by a straight line equal to GU, inscribed in the *"^^ circle BCDE: Let this be the circuraference KB: Therefore the straight line KB is less than GU: And because the angle BZK is obtuse, as was proved in the preceding, therefore BK is greater than BZ : But GU is greater than BK ; much more then is GU greater than BZ, and the square of GU than tbe square of BZ j and AU is equal to AB ; therefore the square of AU, that is, the squares of AG, GU, are equal to the square of AB, that is, to the squares of AZ, ZB ; but the square of BZ is less than the square ofGU ; therefore the square of AZ is greater than the square of AG, and the straight line AZ consequently greater than the straight line AG. Cor. And if in the lesser sphere there be described a solid polyhedron, by drawing straight lines betwixt the points in which the straight lines from the centre of the sphere drawn to all the angles of the solid polyhedron in the greater sphere meet the superficies of the lesser ; in the same order in which are joined the points in which the same lines from the centre meet the superficies of the greater sphere ; the solid polyhe- dron in the sphere BCDE has to this other solid polyhedron the triplicate ratio of that which the diameter of the sphere BCDE has to the diameter »f the other sphere : For if these two solids be divided into the same number of pyramids, and in the same order, the pyramids shall be similar to one another, each to each : Because they have the solid angles at their common vertex, the centre of the sphere, the same in each pyramid, and their other solid angle at the bases equal to one another, each to each% because they are contained by three* B.ii. plane angles, each equal to each; and the pyramids are contained by the samenumber of similar planes; and are therefore similar'* "^^'^t^-^^* to one another, each to each : But similar pyramids have to one another the triplicate*^" ratio of their homologous sides. 'Cor.g.ia. Therefore the pyramid of which the base is the quadrilateral KBOS, ami vertex A, has to the pyramid in the other sphere of the same order, the triplicate ratio of their homologous sides, that is, of that ratio which AB from the centre of the greater sphere has to the straight line from the same centre to tae superficies of the lesser sphere. And in like manner, each pyramid in the greater sphere has to each of the same order in the lesser, the triplicate ratio of that which AB has to the semidiameter of the lesser sphere. And as one antecedent is to its consequent, so are all the antecedents to all the consequents. Wherefore the whole solid polyhedron in the greater sphere has to the whole solid polyhedron in the other, the triplicate ratio of he 284 THE ELEMENTS nc B(x>K XII. of that which AB the semidiameter of the first has to t q^^- ^^''^"'^ midiameter of the other ; that is, which the diameter B 0^ the greater has to the diameter of the other sphere. » n. 12. PROP. XVIII. THEOR. oPHERES have to one anotlier the triplicate ratio of that which their diameters have. Let ABC, DEF be two spheres, of which the diameters are BC, EF. The sphere ABC has to the sphere DEF the tripHcate ratio of that which BC has to EF. For, if it has not, the sphere ABC shall have to a sphere either less ®r greater, than DEF, the triplicate ratio of that which BC has to EF. First, let it have that ratio to a less, viz. to the sphere GHK ; and let the spliere DEF have the same centre with GHK j and in the greater sphere DEF describe" a solid polyhedron, the superficies of which does not meet the lesser sphere GHK ; and in the sphere ABC describe another similar to that in the sphere DEF : Therefore the solid poly- hedron in the sphere ABC has to the solid polyhedron in the Cor. 17. sphere DEF, the triplicate ratio^ of that which BC has to EF. But the sphere ABC has to the sphere GHK, the triplicate ratio of that which BChas toEF; therefore as the sphere ABC to the sphere GH K, so is the said polyhedron in ihcsphere ABC to the solid polyhedron in the sphere DEF : But the sphere = 14.5. ABC is greater than the solid polyhedron in it; therefore* also the sphere GHK is greater than thesolid polyhedron in the sphere DEF : But it is also less, because it is contained within it, which is impossible ; Therefore the sphere ABC has not to any OF KUCLID. 285 any sphere less than DEF, the triplicate ratio of that which ^<^^ xir. BC has to EF. In the same manner, it may be demonstrated, '"^^^'^^ that the sphere DEF has not to any sphere less than ABC, the triplicate ratio of that which EF has to BC. Nor can the sphere ABC have to any sphere greater than DEF, the tripli- cate ratio of that which Be has to EF : For, if it can, Jet it have that ratio to a greater sphere LiVlN : Therefore, by in- version, the sphere LMN has to the sphere ABC, the tripli- cate ratio of that which the diameter EF has to the diameter BC. But as the sphere LMN to ABC, so is the sphere DEF to some sphere, which must be less'^ than the sphere ABC, because the sphere LMN is greater than the sphere DEF : therefore the sphere DEF has to a sphere less than ABC the triplicate ratio of that which EFhas toBC ; which was shewn to be impossible ; Therefore the sphere ABC has not to any sphere greater than DEF the triplicate ratio of that which BC has to EF : and it was demonstrated, that neither has it that ratio to any sphere less than DEF. Tnerefore the sphere ABC has to the sphere DEF, the triplicate ratio of that which BC has to EF. Q. E. D. END OF THE ELEMENTS. I NOTES, CRITICAL AND GEOMETRICAL ; CONTAINING An Account of those Things in which this Edition differs from the Greek Text ; and the Reasons of the Alterations which have been made. As also Observations on some of the Propositions. By ROBERT SIMSON, M. D. Emeritus Professor of Mathematics in the University of Glasgow, LONDON: Printed for F. Wingrave, in the Strand, Successor to Mr. Noubse, 1806. NOTES, &c. aS9 DEFINITION I. BOOK I. It is necessary to consider a solid, thatis, a magnitude which Book i. has length, breadth, and thickness, in order to understand *-^v"*' aright the definitions of a point, line, and superfices ; for these all arise from a solid, and exist in it : The boundary, or boun- daries which contain a solid are called superficies, or the boun- dary which is common to two solids which are contiguous, or which divides one solid into two contiguous parts, is called a superficies : Thus, if BCGF be one of the boundaries which contain the solid ABCDEFGH, or which is the common boundaryofth.ssoiid,andtheso|id BKLCFNMG,and isthere- forein the one as well as in the other solid, is called a superficies, and has no thickness : For, if it have any, this thickness must either be; a part of the thickness H G ,./ k/ NT / B C x/ / / M .\ B of the solid AG, or of the solid BM, or a part of the thickness of each of them. It cannot be a part of the thickness of the solid BM ; because if this solid be removed from the solid AG, the superficies BCGF, the boundary of the solid AG, remain still the same as it was. Nor can it be a part of the thickness of the solid AG j because if this be re- moved from the solid BM, the superficies BCGF, the boundary of the solid BM does nevertheless remain, therefore the super- fices BCGF has no thickness, but only length and breadth. The boundary of a superficies is called a line, or a line is the common boundary of two superficies that are contiguous, or which divides one superficies into two contiguous parts : Thus if BC be one of the boundaries which contain the superficies ABCD, or which is the common boundary of this superficies, and of the superficies KBCL which is contiguous to it, this boundary BC is called a line, and has no breadth : For if it have any, this must be part either of the breadth of the super- ficies ABCD, or of the' superfices KBCL, or part of each of them. It is not part of the breadth of the superfices KBCL ; for, if this superfices be removed from the superfices ABCD, U th(; 290 Book I. ■/ f/ V / T> C / / / A. B K NOTES. the line BC, which is the boundary of the supeifices ABCD, remains the same as it was: Nor can the breadth that BC is supposed to have, be a part of the breadth of the superficies ABCD J because, if this be removed from the superficies KBCL, the line BC, which is the boundary of the superficies KBCL, d#>es nevertheless remain : Therefore the line BC has no breadth : And because the line BC is in a superficies, and that a superficies has no thickness, as was shewn, therefore a line has raither breadth nor thickness, but only length. Tlie boundary of a line is called a point, or a point is the common boundary or extremity of two lines that are contiguous : 1\ G 1^1 Thus, if B be the extremity of the line AB, or the commoii ex- tremity of the two lines AB, KB, this extremity is called a point, and has no length : For if it have any, this length must either be part of the length of the line AB, or of the line KB. It is not part of the length of KB ; for if the line KB be removed from AB, the po1nt"^B which is the extremity of the line AB remains the same as it was : Nor is it part of the length of the line AB ; for, if AB be removed from the line KB, the point B, which is the extremity of the line KB, does nevertheless remain : rherefore the point B has nolength : And because a point js in a line, and a line has neither breadth nor thickness, therefore a point has no length, breadth, nor thickness. And in this manner the defi- nitions of a point, line, and superficies, are to be understood. DEF. VII. B. I. Instead of this definition as it is in the Greek copies, a more distinct one is given from a property of a plane super- ficies, which is manifestly supposed in thelElements, viz. that a st/aight line drawn from any point in a plane to any other in it, is wholly in that plane. DEF. VIII. B. I. It seems that he who made this definition designed that it should comprehend not only a plane angle contatned by two straight lines, but likewise the angle which some conceive to be made by a straight Ime and a curve, or by two curve lines ( which meet one another in a plane : But, though the meaning of the NOTES. 291 the words c-n Ey^£»2f, that is, in a straight line, or in the same BjokL direction, be plain, when two straight lines are said to be in a ^"^' straight line, it does not appear what ought to be understood by these words, when a straight line and a curve, or two curve lines, are said to be in the same direction ; at least it cannot be explained in this place ; which makes it probable that this de- iinition, and that of the angle of a segment, and what is said of the angle of a semicircle, and the angles of segments, in the i6thand 31st Propositions of Book 3, are the additions of some less skilful editor : On which account, especially since thejr are quite useless, these definitions are distinguished from the rest by inverted double commas. DEF. X\ai. B. I. ■ The words, " which also divides the circle into two equal *' parts" are added at the end of this definition in all the copies, but are now Irft out as not belonging to the definition, being only a corollory from ic. Procius demonstrates it by conceiv- ing one of the parts into which the diameter divides the circle, to be applied to the other j for it is plain they must coincide, else the straight lines from the centre, to the circumference would not be all equal : The same thing is easily deduced from the 31st Prop, of Book 3, and the 24ch of the same; from the first of which it follows, that semicircles are similar segments of a Circle ; and from the other, that they are equal to one another. DEF. XXXIII. B. I. This definition has one condition more than is necessary; because every quadrilateral figure which has its opposite sides equal to one another, has likewise its opposite angles equal ; and on the contrary. Let ABCD be a quadrilateral figure, of which the opposite sides AB, CD, are equal to one ano- ther ; as also AD and BC : Join BD i the two sides AD, DB are equal to the two CB, BD, and the base AB is equal to the base CD ; therefore, by Prop. 8. of Book I. the angle ADB is .equal to the an^le CBD; and, by Prop. 4. B. i. the angle BAD is equal ^to the angle DCB, and ABD to BDC ; and therefore also the angle ADC is equal to the angle A^C. U X And 292 N O T E S. ^ooK^ And if the angle BAD be equal to the opposite angle BCD, '*^^^^''*^ and the angle ABC to ADC; the opposite sides are equal: Because, by Prop. 32. B: i. all the angles of the quadrilateral figure ABCD are together equal to four right angles, and the two angles BAD, ADC are together equal to the two angles BCD, ABC : Where- fore BAD, ADC are the half of all the four angles ; that is, BAD and " ADC are equal to two right angles : and therefore AP, CD are parallels by Prop. 28. B. i. In the same manner, AD, BC are parallels : Therefore ABCD is a parallelogram, and its opposite sides are equal, by 34th Prop, B. 1. PROP. VII. B. I. There are two cases of this proposition, one of which is not in the Greek text, but is as necessary as the other : And that the case left out has been formerly in the text, appears ■ plainly from this, that the second part of Prop. 5. which is necessary to the demonstration of this case, can be of no use at all in the Elements, or any where else, but in this demonstra- tion ; because the second part of Prop. 5. clearly follows from the first part, and Prop. 13. B. i. This part must therefore have been added to Prop. 5, upon account of some proposition betwixt the 5th and 13th, but none of these stand in need of it except the 7 th Proposition, on account of which it has been added : Besides, the translation from the Arabic has this case explicitly demonstrated. And Proclus acknowledges, that the second Part of Prop. 5. was added upon account of Prop. 7. but gives a ridiculous reason for it, " that it might afford an " answer to objections made against the 7th," as if the case of the 7th, which is left out, were, as he expressly makes it, an objection against the proposition itself- Whoever is curious may read what Proclus says of this in his commentary on the 5th and 7th Propositions ; for it is not worth while to relate his trifles at full length. , ► . It was thought proper to change the enunciation of this jth Prop, so as to preserve the very same meaning ; the literal translation from the Greek being extremely harsh, and diffi- cult to be understoud by beginners. NOTES. PROP. XL B. I. A COROLLARY IS added to this proposition, which is ne- cessary to Prop. I. B. II. and otherwise. PROP. XX. and XXI. B. I. Proclus, in his commentary, relates, that the Epicureans derided this proposition, as being manifest even to asses, zni needing no demonstration; and his answer is, that though the truth of it be manifest to our senses, yet it is science which must give the reason why two sides of a triangle are greater than the third : But the right answer to this obje<Slion against this and the 21st, and some other plain propositions, is, that the number of axioms ought not to be increased without ne- cessity, as it must be if these propositions be not demonstrated. Mons. Clairauit, in the Preface to his Elements of Geometry, published in French at Paris, anno 1741, says. That Euclid has been at the pains to prove, that the two sides of a triangle which is included within another, are together less than the two sides of the triangle which includes it ; but he has forgot to add this condition, viz. that the triangks must be upon the same base ; because, unless this be added, the sides of the in- cluded triangle may be greater than the sides of the triangle which includes it, in any ratio which is less than that of two to one, as Pappus Alexandrinus has demonstrated in Prop. 3. B. 3. of his mathematical collections. 293 Book I. PROP. XXII. B. I. Some authors blame Euclid because he does not demonstrate ir.it the two circles made use of in the construction of this problem must cut one another : But this is very plain from the determination he has given, viz. that any two of the straight lines DF, FG, GH must be greater than the third : For who is so dull, though only beginning to learn the Elements, as not to perceive that the circle describ- ed from the centre F, at the distance FD, must meet FH betwixt F and H, because FD is less than FH ; and that, , for the like reason, the circle de- scribed from the centre G, at the distance GH or GM, must U 3 meet 294 NOTES. ,^°° K I. meet DG betwixt D and G ; and that these circles must meet ^ one another, because FD and GH are together greater than FG? And this determination is easier to be understood than that which Mr. Thomas Simpson derives from it, and puts instead __^ of Euclid's, in the 49th page of ]5~M Y G H his Elements of Geometry, that he may supply the omission he blames Euclid for, which de- termination is, that any of the three straight lines must be less than the sum, but greater than the difference of the other two: From this he shews the circles must meet one another, in one case ; and says, that it may be proved after the same manner in any other case ; But the straight line GiVI, v/hich he bids take ■^ from GF may be greater than it, as in the figure here annexed ; in which case his dem.onstration must be changed into another. PROP. XXIV. B. I. To this is added, " of the two sides DE, DF, let DE be ** that which is not greater than the other j" that is, take that side of the twoDE, DF which is not greater than the other, in order to make with It the angle EDG jy equal to BAG; because without this restriction there mig-ht be three diffe- rent cases of the proposition, as Cam- panus and others make. Mr. Thomas Simpson, in p. 262 of the second edition of his Elements of Geometry, printed anno 1760, ob- serves in his notes, that it ought to have been shewn, that the points F fall below the line EG. This proba- bly Euclid omitted, as it is very easy to perceive, that DG being equal to DF, the point G is in the circumference of a circle described from the centre D at the distance DF,and must be in that part of it which is above the straight line EF, be- cause DG falls above DF, the angle EDG being greater than the angle EDF. PROP. XXIX. B. I. The propositiort which isusuallycalled the 5th postulate, or jith axiom, by some the i2th,on which this 29th depends, has given NOTES. 295 g-iven a great deal to do, both to ancient and modern geometers : Boo k i. It seecns not to be properly placed among the a-xioms, as indeed '^^^•^^^ it is not self-evident ; but it may be demonstrated thus : DEFINITION I. The distance of a point from a straight line, is the perpen- dicular drawn to it from the point. DEF. 2. One straight line is said to go nearer to, or further from, another straight line, when the distances of the points of the first from the other straight line become less or greater than they were ; and two straight lines are said to keep the same distance from one another, when the distance of the points of one of them from the other is always the same. AXIOxM. A STRAIGHT line cannot first come nearer to another straight line, and then go further from it, before it cuts it; and, in like manner, a straight line cannot go further fr^m another straight line,and then come nearer to it; nor can a straight line keep the same distance from another straight line, and then come nearer to it, or go further from it; for a straight line keeps always the same direction. For example, the straight line ABC cannot first come nearer to the straight line DE,as from the point A to the point B, and then, ^ g See the from the point B to the point C, go ^') ^ above further from the same DE : And, in _ ____^ E ^s«fe- like manner, the straight line FGH ^ G H cannot go further from DE, as from F to G, and then, from G to H, come nearer to the same DE : And so in the last case, as in fig. 2. PROP. I. If two equal straight lines AC, BD, be each at right angles to the same straight line AB : If the points C, D be joined by the straight line CD, the straight line EF drawn from any point E in AB unto CD, at right angles to AB,'shall be equal to AC, or BD. If EF be not equal to AC, one of them must be greater than the other ; let AC be the greater j then, because FE is U 4 less 296 NOTES. = 4. 1. s. 1. BookJL^ less than CA, the straight line CFD is nearer to the straight line AB at the point F than at the point C, that is, CF comes nearer to AB from the point C to F : But because DB is greater than FE, the straight line CFD is further from AB at the point D than at F, that is, FD goes further from AB from F to D : Therefore the straight line CFD first comes nearer to the straight line AB, and then goes further from it, before it cuts it ; which is impossible. Jf FE be said to be greater than CA, or DS, the straight line CFD first goes fur- ther from the straight line AB, and then comes nearer to it: which is also impossible. Therefore FE is not unequal to AC, that is, it is equal to it. PROP. II. If two equal straight lines AC, BD be each at right angles to the same straight lineAB; the straight line CD which joins their extremities makes right angles with AC and BD. Join AD, BCj and because, in the triangles CAB, DBA, CA, AB are equal to DB, BA, and the angle CAB equal to the angle DBA ; the base BC is equal » to the base AD : And in the triangles ACD, BDC, AC, CD are equal to BD, DC, and the base AD is equal to the base BC : Therefore the angle ACD is equaP to the angle BDC : From any point E in AB draw EF unto CD, at right angles to AB ; there- fore, by Prop. I. EF is equal to AC, or BD ; wherefore, as has been just now shewn, the angle, ACF is equal to the angle EFC : In the same manner IS the angle BDF equal to the angle EFD ; but the angles ACD, BDC are equal ; 10. def. 1. therefore the angles EFC and EFD arc equal, and right angles^ j wherefore also the angles ACD, BDC are right angles. CoR. Hence, if two straight lines AB, CD be at right an- gles to the same straight line AC, and if betwixt them a straight line BD be drawn at right angles to cither of them, as to AB; then BD is equal to AC, and BDC is a right angle. If AC be not equal to BD, take BG equal to AC, and join CG: Therefore, by this proposition, the angle ACG is a right anglpi but ACD is also a right angle j wherefore the angles ACD, 297 Book I. N O T £ S. ACD, ACG are equal to one another, whnch is impassible. Therefore BD is equal to AC j and by this proposition BDC "'^^'" is a right angle. PROP. 3. If two straight lines which contain an angle be produced, there may be found in either of them a point from which the perpendicular drawn to the other shall be greater than any given straight line. Let AB, AC be two str^ght lines which make an angle with one another, and let AD hi the given straight line; a point may be found either in ABor AC, as in AC, from which the perpendicular drawn to the other AB shall be greater than AD. In AC take any point E, and draw EF perpendicular to AB ; produce AE to G, so that EG be equal to AE j and produce FE to H, and make EH equal to FE, and join HG. Because, in the triangles AEF, GEH, AE, E!F are equal to GE, EH, each to each, and contain equal * angles, the angle » 15 j_ GHE is therefore equal '' to the an::le AFE which is a right ^^ ( angle : Draw GK perpendicular to AB ; and because the straight lines FK, HG are at right an- ^ ^^ t -d ,- glestoFH, and .p ^'- ^ ^ ^^^ KG at right an- "■'" *- gles to FK, KG is equal to FH, by Cor. Pr. 2. that is, to the double of FE. In the same manner if AG be produced to L, so that GL be equal to AG, and LM be drawn perpendicular to AB, then LM is double of GK, and so on. In AD take A\ equal to FE, and AO, equal to KG, that is, to the double ofFE, or AN; also, take AP, equal to LM, that is, to the double of KG, or AO ; and let this be done till the straight line taken be greater than AD : Let this straight line so taken be AP, and because AP is equal to LM, therefore LM is greater than AD. Which was to be done. PROP. 4. If two straight lines AB, CD make equal angles EAB, ECD with another straight line EAC towards the same parts of it; AB and CD are at right angles to some straight line. Bisect Book I. » 15. I. "4. 1. 298 N O T E S. ^ Biseel: AC in F, and draw ¥G perpendicular to AB ; take CH in the straight ]ine CD, equal to AG, and on the contrary side of AC to that on which AG is, and join FH : Theretbre, in the triangles AFG, CFH, the sides FA, AG are equal to FC, CH, each to each, and the angle FAG, that^ is EAB, is equal to the an^le FCH ; wherefore*' the angle AGF is equal to CHF, aiui AFG to the angle CFH: To these last add the common angle AFH -, therefore the two angles AF"G, AFH are equal- to the two angles CFH, HFA, which two last are equal together to • \^. i. two right angles*^ : therefore also * li. 1, AFG, AFH are equal to two right angles, and consequently** GF and FH are in one straight line. And because AGF is a right angle, CHF which is equal to it is also a right angle ; Therefore the straight lines AB, CD are at rightangles to GH. C H I) 2C\ I. •> 13. 1. PROP. 5. If two straight lines BB, CD be cut by a third ACE, so as to make the interior angles BAC, ACD, on the same side of it, together less than two right angles; AB and CD being produced, shall meet one another towards the parts on which are the two angles, which are less than two right angles. At the point C, in the straight line CF, make^ the angle ECF equal to the angle EAB, and draw to AB the straight line CG at right angles to CF : Fhen, because the angles ECF, E AB^are equal to one an- other, and that the angles K ECF, B'CA are together equal*" to two right an- gle's, the angles EAB, FCA are equal to two right angles. But by the hypothesis, the angles EAB, ACD are toge- ther less than two right angles ; therefore theangle FCA is greater than ACD, and CD falls between CF and AB : And because CF and CD make an angle with one ano- tber, by Prop. 3. a point may be found in either of them CD^ from which the perpendic-ular drawn to CF shall be greatCj. N O T E S. 299 than tlie sfraight line CG. Let thrs point be H, and dtaw ,^^^^J\ HK perpendicular to CF, meeting AB in L : And because AB, CF contain equal angles with AC on the same side of it, by Prop. 4. AB and CK are at right angles to the straight line MNO, which bisects AC in N,and is perpendicular to CF ; Therefore by'Cor. Prop. 2. CGand KL, which are at right angles to CF, are equal to one another : And HK is greater than CG, and therefore is greater than KL, and consequently the point H is in KL produced. Wherefore the straight line CDH, drawn betwixt the points C, H, which are on contrary sides of AL, must necessarily cut the straight line AB. PROP. XXXV. B. T. The demonstration of this Proposition is changed, because, if the method which is used in it was followed, thtrre would be three cases to be separately demonstrated, as is done in the translation from the Arabic; for, in the Elements, no case of a Proposition that requires a different demonstration ought"tp be omitted. On this account, we have chosen the method which Morfs. Ciairault has given, the first of any, as far as I know, in his Elements, page 21, and which afterwards Mr. Simpson gives in his page 32. But whereas Mr. Simpson makes use of Prop. 26. B. i. from which the eq'jality of the two triangles does not immediately follow, because, to prove that, the 4th of B. 1. must likewise be»made use of, as may be seen in the very same case in the 34th Prop. B. i. it was thought better to make use only of the 4th of B. i. PROP. XLV. B.I. The straight line KM is proved to be parallel to FI, from the 33d Prop, whereas KH is parallel to FGby contruction, and KHAl, FGL have been demonstrated to be ^-.traight lines. A corollary is added from Commandine, as being often used. PROP. xiir. B. II. J.N this proposition only acute angled triangles are men- BookTL tioned, whereas it holds true of every triangle: and the demon- ^•^s'^^ strations of the cases omitted are added; Commandine andCla- vius have likewise given their demonstrations of these cases. PROP. XIV. B. II. In the demonstration of this, some Greek editor lias igno- rantly inserted the words, " but if not, one of the two BE, " ED, NOTES. " ED, is the greater t Let BE Ijc the greater, and produce it to " F," as if it was of any consequence whether the greater or lesser be produced : Therefore, instead of the:?e words, there ought to be read only, *' but if not, produce BE to F." Book III. Q PROP. I. B. Ill, EVERAL authors, especially among the modern mathe- maticians' and logicians, inveigh too severely against indirect or apogogic demonstrations, and sometimes ignorantly enough; not bemg aware that there are some things that cannot be de- monstrated any other way : Of this thu present proposition is a very clear instance, as no direct demonstration can be given of it : Because, besides the definition of a circle, there is no principle or property relating to a circle antecedent to this problem, from which either a direct or indirect demon- stration can be deduced : Wherefore it is necessary that the point found by the construction of the problem be proved to be the centre of the circle, by the help of this definition, and some of the preceding propositions : And because, in the de- monstration, this proposition must be brought in, viz. straight lines from the centre of a circle to the circumference are equal, and that the point found by the construction cannot be as- sumed as the centre, for. this is the thing to be demonstrated : it is manifest some other point must be assumed as the centre ; and if from this assumption an absurdity follows, as Euclid de- monstrates there must, then it is not true that the point as- sumed is the centre; and as any point whatever was assumed, it follows that no point, except that found by the construction, can be the centre, from which the necessity of an indirect demonstration in this case is evident. PROP. XIII. B. III. As it is much easier to imagine that two circles may touch one another within in more points than one, upon the same side, than upon opposite sides ; the figure of that case ought not to have been omitted ; but the construction in the Greek text would not have suited with this figure so well, because the centres of the circles must have been placed near to the cir- cumferences; on which account another construction and demonstration is given, which is the same with the second part of that which Campanus has translated from the Arabic, where, NOTES 3^1 where, without any reason, the demonstration is divided into ^0°^^- two parts. PROP. XV. B. III. Th£ converse of the second part of this proposition is want- ing, though in the preceding, the converse is added, in a like case, both in the enunciation and demomtration; and it is now added in this. Besides, in the demonstration of the first part of this I5thj tiiC diameter AD (see Commandine's figure), is proved to be greater than the straight iine BC by means of another straight line MN ; whereas it maybe better done v/ith- out it : on which accounts we have given a different demon- stration, like to that which Euclid gives m the preceding 14th, and to that which Theodosius gives in Prop. 6. B. i. of his Spherics, in this very affair. PRO?. XVI. B. III. In this we have not followed the Greek r.or the Latin trans- lation literally, but have given what is plainly the meaning of the proposition, without mentioning the angle ofthe semicircle, or that which some call the cornicular angb, which they con- ceive to be made by the circumference and the straight line which is at right angles to the diameter, at its extremity ; which angles have furnished matter of great debate between some ofthe modem gcometrers, and given occasion of deducing strange consequences from them, which are quite avoided by the manner in which we have expressed the proposition. And in like manner, we have given the true meaning of prop. 31. B. 3. without mentioning the angles ofthe greater or lesser segments. These passages Vieta, with good reason, suspetfls to be adulterated in the 386th page of his Oper. Math. . PROP. XX. B. III. The first words of the second part of this demonstration, '' xixXaffS'a/ '8r,zja'>.L\" are wrong translated by Mr. Briggs and Dr. Gregory, " Rursusinclinetur;" for the translation ought to be " Rursus infleclatur," as Commandine has it : A straight line is said to be inflected either to a straight, or curve line, when a straight line is drawn to this line from a point, and from the point in which it meets it, a straight line making an angle with the former is drawn to another point, as is evi- dent from the qoth prop, of Euclid's Data : For thus the whole line betwixt the first and last points is infledtd or broken at the 302 NOTE S. Book in. the point of infle£lion, where the two straight lines meet. And in the like sense two straight lines are said to be infledled from two points to a third point, when they make an angle at this point; as may be seen in the description given by Pappus Alexandrinus of Appollonius's Books de Locis planis, in the preface to his yth Book : We have made the expression fuller from the 90th Prop, ot the Data. ' PROP. XXr. B. III. There are two cases of this proposition, the second of which, viz. when the angles are in a segment not greater than a semicircle, is wanting in the Greek : And of this a more simple demonstration is given than that which is in Comman- dine, as being derived only from the first case, without the help of triangles. PROP. XXIIl. and XXIV. B. III. In proposition 24. it is demonstrated that the segment AEB must coincide with the segment CFD (see Comman- dine's figure,) and that it cannot fail otherwise, as CGD, so as to cut the other circle in a third point G, from this, that, if it did, a circle could cut another in more points than two: But this ought to have been proved to be impossible in the 23d prop, as well as that one of the segments cannot fall '^'ithiri the other. This part, then, is left out in the 24th, and put in its proper place, the 23rd proposition. PROP. XXV. B. III. This proposition is divided into three cases, of which two have the same construilion and demoiistration j therefore it is now divided only into two cases. PROP. XXXIII. B. III. This also in the Greek is divided into three cases, o» which, two, viz. one, in which the given angle is acute, and the other in which it is obtuse, have exactly the same con- strudlion and demonstration : on which account, the demon- stration of the last case is left out, as quite superfluous, and the addition of some unskilful editor j besides the demonstra- tion of the case when the angle given is a right angle, is done a roundirabout way, and is therefore changed to a more sim- ple one,- as was done by Clavius. NOTES. PROP. XXXV. B. 111. As the 25rh and 33'-<i propositions are divided into more oases, so this 35th is divided into fewer cases than are necessa- ry. Nof can It be supposed that Euclid omitted them because they are easy ; as he has given the case, which by far is the easiest of them all, viz. that in which both the straight lines pass through the centre: And in the following proposition he separately demonstrates the case in which the straight line passes through the centre, and that in which it does not pass through the centre ; So that it seems Theon, or some other, has thought them too long to insert : But cases that require different demonstrations, should not be leftout in the Elements, a^ was before taken notice of: These cases are in the transla- tion from the Arabic, and are now put into the text. PROP. XXXVII. B. III. At the end of this, the words " in the same manner it may " be demonstrated, if the centre be in AC," are left out as the addition of some ignorant editor. w, DEFINITIONS of BOOK IV. HEN a point is in a straight line, or any other line, this Book IV point is by the Greek geometers said arareffS-ai, to be upon, "-^^^"^ or in that line, and when a straight line or circle meets a cir- cle any way, the one is said u^rsa^ai to meet the other : But when a straight line or circle meets a circle so as not to cut it, it is said g^pasTTSffS-aj, to touch the circle ; aqd these two terms are never promiscuously used by them : Therefore, in the 5th definition of B. 4. the compound £{paiSTr/T«imustbe read, instead of the simple aSJTy/Tsci : And in the ist, 2d, 3d, and 6th defi- nitions in Commandine's translation, " tangit," must be read instead of" contingit :" And in the 2d and 3d definitions of Book 3. the same change must be made : But in the Greek text of propositions nth, 12th, 13th, i8th, 19th, Book 3. the compound verb is to be put for the simple. PROP. IV. B. IV. In this, as also in the 8th and I3ch proposition of this book, it is demonstrated indirectly, that the circle touches a straight line; whereas in the 17th, 33d, and 37th propositions of Book 3. the same thing is directly demonsti'ated : And this way we have 394 NOTES. Book IV. havc chosci) to usc in the propositions of this book, as it is ""^^^'"^ shorter. PROP. V. B. IV. The demonstration of this has been spoiled by some un- skilful hand: For he does not demonstrate, as is necessary, that the two straight lines which bisedls the sides of the tri- angle at right angles must meet one another ; and, without any reason, he divides the proposition into three cases j whereas, one and the same construdfion and demonstration serves for them all, as Campanus has observed ; which useless repetitions are now left out : The Greek text also in the corollary is manifestly vitiated, where mention is made of a given angle, though there neither is, nor can be, any thing in the proposi- tion relating to a given angle. PROP. XV. and XVI. B. IV. In the corollary of the first of these, the words equilateral and equiangular are wanting in the Greek : and in prop. i6. instead of the circle ABCD, ought to be read the circumfe- rence ABCD : Where mention is made of its containing fif- teen equal parts. DEF. III. B. V. JooK V jLVi-ANY of the modern mathematicians rejeft this defini- '"^^'^'^ tion : The very learned Dr. Barrow has explained it at large at the end of his third lefture of the year 1666, in which also he answers the objections made against it as well as the sub- ject would allow : And at the end gives his opinion upon the whole, as follows ; " I shall only add, that the author had, perhaps, no other ** design in making this definition, than (that he might more *' fully explain and embellish his subjecl) to give a general *' and summary idea of ratio to beginners, by premising " this metaphysical definition, to the more accurate defini- *' tioiis of ratios that are the same to one another, or one of *^ which is greater, or less than the other: I call it a meta- " physical, for it is not properly a mathematical, definition, *' since nothing in mathematics depends on it, or is deduced, *' nor, as I judge, can be deduced from it : And the defini- ** tion of analogy, which follows, \\z. Analogy is the simi- 3 • " litudc NOTES. 3»5 " litude of ratios, is of the same kind, and can serve for no B^ok v. *' purpose in mathematics, but only to give beginners some ge- ^"^""^^ ** neral, though gross and confused, notion of analogy: But the ** whole of the doilrine of ratios, and the whole of mathema- *' ticsjdependupontheaccurate mathematical definitions which *' tallow this : To these we ought principally to attend, as the ** doctrine of ratios is more perfectly explained by them j this *' third, and others like it, may be entirely spared without any *' loss to geomecry ; as we see in the 7th book of the Elements, *' where the proportion of numbers to one another is defined, ** and treated of, yet without giving any definition of the ratio " of numDers ; though such a definition was as necessary and *' useful to be given in that hookas in this : But indeed there *' is scarce any need of it in either of them : Though I think ' ' that a thing of so general and abstracted a nature, and there- *' by the more difficult to be conceived and explained, cannot *' be more commodiously defined than as the author has done : *' Upon which account I thought fit to explain it at large, and " defend it against the captious objections of those who attack *' it." To this citation from Dr. Barrow I have nothing to add, except that I fully believe the 3d and 8th definitions are not Euclid' S) but added by some unskilful editor. DEF. XL B. V. It was necessary to add the word " continual" before " proportionals" in this definition; and thus it is cited in the 33d prop, of Book II. After this definition ought to have followed the definition of compound ratio, as this was the proper place for it; duplicate and triplicate ratio being species of compound ratio: ButTheon has made it the 5th def. of B. 6. where he gives an absurd and entirely useless definition of compound ratio : For this reason we have placed another definition of it betwixt the nth and I2th of this book, which, no doubt, Euclid gave; for he cites it expressly in Prop. 23. B. b, and which Clavius, Herigon, and Barrow, have likewise given, but they retain also Theon's, which they ought to have left out of the Elements. DEF. XIII. B. V. This, and the rest of the definitions following, contain the explication of some terms which are used in the 5th and foj.. lowing books; which, except a icw, are easily enough under* X stood 3o6 NOTES. Book V. Stood from the propositions of this book where they are first mentioned : They seem to have been added by I'heon, or some other. However it be, they are explained something more distinctly for the sake of learners. PROP, IV. B. V. In the construftion preceding the demonstration of this, the words » £Tyx,£) any wliatever, are twice wanting in the Greek, as also in the Latin translations j and are now added, as being wholly necessary. Ibid, in the demonstration ; in the Greek, and in the Latin translation of Commandine, and in that of Mr. Henry Briggs, which was published at London in 1620, together with the Greek text of the first six bocks, which translation in this place is fallowed by Dr. Gregory in his edition of Euclid, there is this sentence following, viz. " and of A and C have been taken ." equimultiples K, L ; and of B and D, any equimultiples " whatever (« ct^xO ^i ^ ;" which is not true, the words " any whatever," ought to be left out : And it is Strange that neither Mr. Briggs, who did right to leave out these words in one place of Prop. 13, of this book, nor Dr. Gregory, who changed them into the word *•' some" in three places, and left them out in a fourth of that same Prop. 13, did not also leave them out in this place of Prop. 4, and in the second of the two places where they occur in Prop. 17 of this book, in neither of which they can stand consistent with truth : And in none of all these places, even in those whigh they corre<Sted in their Latin translation, have they cancelled the words a Ert/j^e in the Greek text, as they ought to have done. The same words « srt/x,? arc foundinfour places of Prop. 11 of this book, in the first and last of which they are necessary, but in the second and third, though they are true, they are quite superfluous ; as they likewise are in the second of the two places in which they are found in the 12th Prop, and in the like places of Prop. 22, 23, of this book ; but are wanting in the last place of Prop. 23, as also in Prop. 25, Book 11. COR. PROP. IV. B. V. This Corollary has been unskilfully annexed to this propo- sition, afid has been made instead of the legitimate demon- stration, which, without doubt, Theon, or some other editor, has taken away, not from this, but from its proper place in this NOTES. 307 this book: The author of it designed to demonstrate, that if Book v. four magnitudes E, G, F, H, be proportionals, they are also ^•^''^^ proportionals inversely ; that is, G is to E, as H to F j which is true ; but the demonstration of it does not in the least depend upon this 4th prop, or its demonstration : For, when he says, *' because it is demonstrated, that if K be greater than M, L is ** greater than N," &c. This indeed is shewn in the demon- stration of the 4th prop, but not from this, that E, G, F, H, are proportionals ; for this last is the conclusion of the propo- sition. Wherefore these words, " because it is demonstrated," &c. are wholly foreign to his design : And he should have proved, that if K be greater than M, L is greater than N, from this, that E, G, F, H, are proportionals, and from the 5th def. of this book, which he has not ; but is done in proposition B, which we have given in its proper place, instead of this co- rollary i and another corollary is placed after the 4th prop, which is often of use ; and is necessary to the demonstration of Prop. 18 of this book. PROP. V. B. V. In the construftion which precedes the demonstration of this proposition, it is required that EB may be the same mul- tiple of CG, that AE is of CF ; that is, that EB be divided into as many equal parts, as there are parts in AE equal to CF : From which it is evident, that this construction is not Euclid's, for he does not shew the way of dividing straight lines, and far less other magnitudes, into any number of equal parts, until the 9th proposition of B. 6. and he nevpr requires any thing to be done in the construction of which he had not before giren the method of doing : For this rea- son, we have changed the construdlion to one. C which, without doubt, is Euclids, in which no- thing is required but to add a magnitude to itself a certain number of times ; and this is to be found in the translation from the Arabic, though the enunciation of the proposition and the demonstra- tion are there very much spoiled. Jacobus Peleta- rius, who was the first, as far as I know, who took B^ J) notice of this error, gives also the right construc- tion in his edition of Euclid, after he had given the other which he blames : He says, he would not leave it out, because it was fine, and might sharpen one's genius to invent others like it ; ' X 2 whereas 3o8 NOTES. Book V, whereas thercis not the least difference between the twodemon- ■^**''"''"*'^ strations, except a single word in the constru^lion, which very probably has been owing to an unskilful librarian. Claviuslike- wise gives both the ways ; but neither he nor Peletarius takes notice of the reason why the one is preferable to the other. PROP. VI. B. V. There are two cases of this proposition, of which only the first and simplest is demonstrated in the Greek : And it is pro- bable Theon thought it was sufficient to give this one, since he was to make use 'of neither of them in his mutilated edition of the 5th book ; and he might as well have lett out the other, ■ as also the fifth proposition, for the same reason. The demon- stration,ot the other case is now added, because both of them, as also the 5th proposition, are necessary to the demonstration of the i8th proposition of this book. The translation from the Arabic gives both cases briefly. PROP. A. B. V. This proposition is frequently used by geometers, and it is necessary in the 25th prop, of this book, 31st of the 6th, and 34th of the nth, and 15th of the 12th book : It seems to have been taken out of the Elements by Theon, because it appeared evident enough to him, and others, who substitute the confused and indistindl idea the vulgar have of proportionals, in place of that accurate idea which is to be got from the 5th def. of this book. Nor can there be any doubt that Eudoxus or Euclid gave it a place in the Elements, when we see the 7th and 9th of the same book demonstrated, though they are quite as easy and evident as this. Alphonsus Borellus takes occasion from this proposition to censure the 5th definition of this book very severely, but most unjustly. In p. 126 of his Euclid Restored, printed at Pisa in 1658, he says, " Nor can even this least ** degree of knowledge be obtained from the foresaid property," viz. that which is contained in 5th def. 5. '' That, if four *' magnitudes be proportionals, the third must necessarily be " greater than the fourth, when the first is greater than the *' second : as Clavius acknowledges in the i6th prop, of the ** 5th book of the elements." But though Clavius makes no such acknowledgment expressly, he has given Borellus a han- dle to say this of him; because -Cvhen Clavius, in the above cited place, censures Commandine, and that very justly, for demonstrating this proposition by help of the 16th of the 5th j yet he himself gives no demonstration of it, but thinks it plain from NOTES. 309 from the nature of proportionals, as he writes in the end of the Book. v. 14th and i6:h Prop. B. 5. of his edition, and is followed by ^^**'**', Herigoa in Schol. I. Prop. 4. B. 5, as if there was any nature of proportionals antecedent to that which is to be derived and understood from the definition of them : And indeed, though it is very easy to give a right demonstration of it, nobody, as far as I know, has given one, except the learned Dr. Barrow, who, in answer to Borellus's objection, demonstrates it indi- rectly, but very briefly and clearly, from the 5th definition, in the 322d page of his Lect. Mathem, from which definition it may also be easily demonstrated directly : On which ac* count we have placed it next to the propositions concerning equimultiples. PROP. B. B. V. This also is easily deduced from the 5th def. B. 5, and therefore is placed next to the other ; for it was very igno- rantly made a corollary from the fourth Prop, of this Book. See the note on that corollary. PROP. C. B. V. This is frequently made use of by geometers, and is neces- sary to the 5th and 6th Propositions of the loth Book. Cla- vius, in his notes subjoined to the 8tli def. of Book 5. demon- strates it only in numbers, by help of some of the propositions of the 7th Book: in order to demonstrate the property con- tained in the 5th definition of the 5ch Book, when applied to numbers, from the property of proportionals contained in the 20th def. of the 7th Book : And most of the commentators judge it difficult to prove that four magnitudes which are pro- portionals according to the 20th def. of 7th Book, are also pro- portionals according to the 5th def. of 5th Book. But this is easily made out as follows : First, if A, B, C, D, be four mag- nitudes, such that A is the same mul- tiple, or the same part of B, vihich C is of D : A, B, C, D, are propor- tionals : This is demonstrated in pro- position C. Secondly, if AB contain the same parts of CD that EF does of GH j in this case likewise AB is to CD, as EF to GH. X3 B D R ACE H L G Let 310 , NOTES. Book V. Let CK be a part of CD, and GL the same part of GH ; ^"**v*^ and let AB be the same multiple of CK, -:hat EF is of GL : Therefore, F by Prop. C, of 5th Book, AB is to B TT CK, as EF to GL : And CD, GH jy are equimultiples^ of CK, GL the se- cond and fourth j wherefore, by Cor. Prop. 4. Book 5. AB is to CD, as EF K to GH. And if four magnitudes be propor- A P E Gl tionals according to the 5th def. of Book 5, they are also proportionals according to the 20th def. of Book 7. ' First, if A be to B, as C to D ; then if A be any multiple or part of B, C is the same multiple or part of D, by Prop. D. of B. 5. Next, if AB be to CD, as EF to GHj then if AB con- tainsany partsofCD, EF contains the same parts of GH: For let CK be a part of CD, and GL the same part of , GH, and let AB be a multiple of CK : EF is the same multiple of GL : take M the same multiple of GL that AB is of CK ; therefore, by Prop. C. of B. 5. AB is to CK, as M toGL : and CD, GH are equimultiples of CK, GL ; wherefore, by Cor. Prop. 4. B. 5. AB is to CD, as M to GH. And, by the hypothesis, AB is to CD, as EF to GH ; therefore M is equal to EF by M Prop. 9. Book 5. and consequently EF is the same multiple of GLthatABisofCK. PROP. D. B. V. This is not unfrequently used in the demonstration of other ' propositions, and is necessary in that of Prop. 9. B. 6. It seems Theon has left it out for the reasons mentioned in the notes at Prop. A. PROP. Viir. B. V. Li the demonstration of this, as it is now in the Greek, there are two cases (see the demonstration in Hervagius, or Dr. Gregory's edition), of which the first is that in which AE is less than EB ; and in this it necessarily follows, that H© the multiple EB is greater than ZH the same multiple of AE, which last multiple, by the construction, is gteater than A j whence also H0 must be greater than A : But in the second case, viz. that in which EB is less than AE, though ZH be greater than A, yet n© may be less than the same A ; so that there NOTES. 3" H A E- HF e B A © there cannot be taken a multiple of a which is the first that is ^°°^ v* greater than K or H0, because A itself is greater than it : ^"^^'^^ Upon this account, the author of this demonstration found it necessary to change one part of the construction that was made use of in the first case : But he has, without any necessity, changed also another part of it, viz. when he orders to tak« N that multiple of a which is the first that is greater than ZH } for he might have taken that multiple of a which is the first that is greater than H©, or K, as was done in the first case: Helikewisebrings in this K into the demonstra- tion of both cases, without any reason; for it serves to no purpose but to lengthen the demonstration. There is also a third case which is not mentioned in this demonstration, viz. that in which AE in the first, or EB in the second of the two other cases, is greater than D ; and in this any equimultiples, as the doubles, of AE, EB are to be taken, as is done in this edition, where all the cases are it once demonstrated : And from this it is plain that Theon, or some other unskilful editor, has vitiated this proposition. PROP. IX. B. V. Of this there is given a more explicit demonstration than that which is now in the Elements. PROP.X. B. V. It was necessary to give another demonstration of this pro- position, because that which is in the Greek and Latin, or other edition?, is not legitimate : For the words greater, the same, or equal, lesser, have a quite different meaning when ap- plied to magnitudes and ratios, as is plain from the 5th and 7th definitions of Book 5. By the help of these let us exa- mine the demonstration ot the loth Prop, which proceeds thus : *' Let A have to C a greater ratio than B to C : I say that A is " greater than B ; for if it is not greater, it is either equal or " less. But A cannot be equal to B, because then each of them *' would have the same ratio to C; but they have not. There- " fore A is not equal to B." The force of whichreasoning is this : If A had to C the same ratio that B has to C, then if X 4 any 3r2 NOTES. Book V. any equimultiples whatever of A and B be taken, and any ^"^'^^^^ multiple whatever of C j if the multiple of A be greater than the multiple of C, then, by the 5th def. of Book 5, the multiple of B is also greater than that of C : but, from the hypothesis that A has a greater ratio to C, than B has to C, there must, by the 7th def. of Book 5, be certain equimultiples of A and B, and some multiple of C such, that the multiple of A is greater than the multiple of C, but the muhiple of B is not greater than the same multiple of C : And this proposition direflly contradi6ls the preceding ; wherefore A is not equal to B. The demonstration of the loth pcop. goes on thus: *' But " neither is A less than B ; because then A would have a less " ratio to C than B has to it : But it has not a less ratio, there- *' fore A is not less than B," &c. Here ijt is said, that " A ** would have a less ratio to C than B has to C," or, which is the same thing, that B would have a greater ratio to C than A to C ; that is, by 7 th def. Book 5, there must be some equi- multiples ef B and A, and some multiple of C, such that the multiple of B is greater than the multiple of C, but the mul- tiple of A is not greater than it : And it ought to have been proved, that this can never happen if the ratio of A to C be greater than the ratio of B to C ; that is, it should have been proved, that, in this case, the multiple of A is always greater than the multiple of C, whenever the multiple of B is greater than the multiple of C j for when this is demonstrated, it will be evident that B cannot have a greater ratio to C, than A has to C, or, which is the same thing, that A cannot have a less ratio to C than B has to C . But this is not at all proved in the loth proposition : But if the lOth were once demonstrated, it would immediately follow from it, but cannot without it be easily demonstrated, as he that tries to do it will find. Where- fore the lOth proposition is not sufficiently demonstrated. And it seems that he who has given the demonstration of the loth proposition as we now have it, instead of that which Eudoxus or Euclid had given, has been deceived in applying what is , manifest, when understood of magnitudes, unto ratios, viz. that a magnitude cannot be both greater and less than another. That those things which are equal to the same are equal ta one another, is a most evident axiom when understood of magnitudes; yet Euclid does not make use of it to infer, that those ratios which are the same to the same ratio, are the same to one another; but explicitly demonstrates this in Prop. ii. of Book 5. The demonstration we have given of the i oth prop. is NOTES. 313 If A have to C a A D E is no doubt the same with that of Eudoxus or Euclid, as it is ^°^^ immediately and directly derived from the definition, of a ^^ greater ratio, viz. the 7 of the 5. The above-mentioned proposition, viz. greater ratio than B to C j and if of A and B there be taicen certain equimulti- ples, and some multiple of C ; then if the multiple of B be greater than the multiple of C, the multiple of A is also greater than the same, is thus demonstrated : LetD, E be equimultiples of A, B, and F a multiple of C, such, that E the mul- tiple of B is greater than F j D the mul- tiple of A is also greater than F. Because A has a greater ratio to C , than B to C, A is greater than B, by the loth Prop. B. 5, therefore D the multiple of A is greater than E the same mulciple of B : And E is greater than F : much more therefore D is greater than F. c PROP. XIII. B. V. In Commandine's, Briggs's, and Gregory's translations, at the beginning of this demonstration, it is said, " And the mul- '' tiple of C is greater than the multiple of Dj but the mul- " tiple of E is not greater than the multiple of F :" Which words are a literal translation from the Greek : But the sense evidently requires that it be read, " so that the multiple of C *' be greater than the multiple of D ; but the multiple of E be "not greater than the multiple of F." And thus this place was restored to the true reading in the first editions of Com- mandine's Euclid, printed in 8vo. at Oxford : But in the later editions, at least in that of 1747, the error of the Greek text was kept in. There is a corollary added to prop. 13, as It Is necessary to the 20th and 21st prop, of this book, and is as useful as the proposition. PROP. XIV. B. V. The two cases of this, ^hlch are not in the Greek, are added ; the demonstration of them not being exactly the same with that of the first case. NOTES. PROP. XVII. B. V. The order of the words in a clause of this Is changed to one more natural : As was also done in pfop, ii. PROP. XVIII. B. V. The demonstration of this is none of Euclid's, nor is it legi- timate ; for it depends upon this hypothesis, that to any three magnitudes, two of which,- at least, are of the same kind, there may be a fourth proportional : which, if not proved, the demonstration now in the text is of no force : But this is assumed without any proof; nor can it, as far as I am able to discern, be demonstrated by the propositions preceding this : so far is it from deserving to be reckoned an axiom, as Cla- vius, after other commentators, would have it, at the end of the definitions of the 5Ch book. Euclid does not demonstrate it, nor does he shew how to find the fourth proportional, before the 12th prop, of the 6th book : And he never assumes any thing in the demonstration of a proposition, which he had not before demonstrated ; at least, he assumes nothing the exist- ence of which is not evidently possible; for a certain conclusion can never be deduced by the means of an uncertain proposition : Upon this account, we have given a legitimate demonstration of this proposition instead of that in the Greek and other edi- tions, which very probably Theon, at least some other, has put in the place of Euclid's, because he thought it too prolix: And as the 17th prop, of which this i8th is the converse, is demonstrated by help of the 1st and 2nd propositions of this book ; so, in the demonstration now given of the 1 8th, the 5th prop, and both cases of the 6th are necessary, and these two propositions are the converses of the ist and 2d. Now the 5th arid 6th do not enter into the demonstrationof any proposition in this book as we now have it : Nor can they be of use in any proposition of the Elements, except in this i8th, and this is a manifest proof, that Euclid made use of them in his demon- stration of it, and that the demonstration now given, which is exactly the converse of that of the 17th, as it ought to be, dif- fers nothing from that of Eudoxus or Euclid : For the 5th and 6th have undoubtedly been put into the 5th book for the sake of some propositions in it, as all the other propositions about equimultiples have been. Hieronymus Saccherius, in his book named " Euclides ab "omni nasvo vindicatus," printed at Milan ann. 1733, in 410, acknowledge s NOTES. 315 acknowledges this blemish in the demonstration of the i8th, ^^^^^^ and that he may remove it, and render the demonstration we now have of it legitimate, he endeavours to denwnstrate the following proposition, which i-s in page 115 of his book, viz. *' Let A, B, C, D be four magnitudes, of which the two " first are of the one kind, and also the two others either of the " same kind with the two first, or of some other the same *' kind with one another. I say the ratio of the third C to the " fourth D, is either equal to, or. greater, or less than the ratio "of the first A to the second B." And after two propositions premised as lemmas, he proceeds thus ; ** Either among all the possible equimultiples of the first A, and of the third C, and, at the same time, among all the possible equimultiples of the second B, and of the fourth D, there can be found some one multiple EF of the first A, and one IK of the second B, that are equal to one another j and also (in the s.cme casQ) some one multiple GH of the third C equal to LM the multiple of the fourth D, or such equality is no where to be found. If the first case happen, [i. e. if such A £ f equality is to be found] it B 1 ^K_ is manifest from what is C G H before demon- strated, that J) J^ J£ A is to B as C to D } but if such simultaneous equality be not to be found upon both sides, it will be found either upon one side, as upon the side of A [and B] ; or it will be found upon neither side j if the first happen : therefore (from Euclid's definition of greater and lesser ratio foregoing) A has to B a greater or less ratio than C to D ; according as GH the multiple of the third C is less, or greater than LM the multiple of the fourth D : But if the second case happen ; therefore upon the one side, as upon the side of A the first and B the sec6nd, it may happen that the multiple EF [viz. of the first] may be less than IK the * multiple of the second, while, on the contrary, upon the, * other side [viz. of C and D], the multiple GH [of the third 'C]is greater than the other multiple LM [of the fourth ** D] : And then (from the same definition of Euclid) the ratio "of 3i6 NOTES. Book V' "of the first A to the second B, is less than the ratio of the ^"^""'f^^ <« third C to the fourth D j or on the contrary. *' Therefore the axiom [i. e. the proposition before set down] "remains demonstrated," &c. Not in the least ; but it still remains undemonstrated j For what he says may happen, may, in innumerable cases, never happen; and therefore his demonstration does not hold: For example, if A be the side, and B the diameter of a square ; and C the side, and D the diameter of another square ; there can in no case be any multiple of A equal to any of B ; nor any one of C equal to one of D, as is well known ; and yet it can never happen that when any multiple of A is greater - than a multiple of B, the multiple of C can be less than the multiple of D, nor when the multiple of A is less than that of B, the multiple of C can be greater than that of D, viz. talcing equimultiples of A and C, and equimultiples of B and D : For A, B, C, D are proportionals j, and so if the multiple of A be greater, &c. than that of B, so must that of C be greater, &c. than that of D ; by 5th Def. b. 5. The same objection holds good against ?he demonstration which some give of the first prop, of the 6th book, which we have made against this of the i8th prop, because it depends upon the same insufficient foundation with the other, PROP. XIX. B. V. A COROLLARY is added to this, which is as frequently used as the proposition itself. The corollary which is subjoined to it in the Greek, plainly shews that the 5th book has been vitiated by editors who were not geometers : For the conversion of ratios does not depend upon this igyth, and the demonstration which several of the commentators on Euclid give of conver- "* sion is not legitimate, as Clavius has rightly observed, who has given a good demonstration of it which we have put i^ proposition E ; but he makes it a corollary from the 19th, and begins it with the words, " Hence it easily follows," though it does not at all follow from it. PROP. XX. XXL XXII. XXIII. XXIV. B. V. The demonstration of the 20th and 21st propositions, are shorter than those Euclid gives of easier propositions, either in the preceding or following books : Wherefore it was pro- per CO make them more explicit, and the 22d and 23d propo- sLons arc, as they ought to bcj extended to any number of magnitudes : NOTES. 317 magnitudes : And, in like manner, may the 24th be, as is Boos V. taken notice of in the corollary j and another corollary is added, ^"^"^"'^'^ as useful as the proposition, and the words, " any whatever" are supplied near the end of prop. 23, which are wanting in the Greek text, and the translations from it. In a paper writ by Philippus Naudaeus, and published after his death, in the history of the Royal Academy of Sciences of Berlin, anno 1745, page 50, the 23d prop, of the 5th book is censured as being obscurely enunciated, and, because of this, prolixly demonstrated : The enunciation there given is not Euclid's, but Tacquet's, as he acknowledges, which, though not so well expressed, is, upon the matter, the same with that which is now in the Elements. Nor is there any thir\g ob- scure in it, though the author of the paper has set down the proportionals in a disadvantageous order, by which it appears to be obscure : But no doubt Euclid enunciated this 23d, as well as the 22d, so as to extend it to any number of magni- tudes, which, taken two and two, are proportionals, and not of six only ; and to this general case, the enunciation which Naudaeus gives, cannot be well applied. The demonstration which is given of this 23d, in that paper, is quite wrong ; because, if the proportional magnitudes be plane or solid figures, there can no rectangle, (which he impro- perly calls a produii) be conceived to be made by any two of them : And if it should be said, that in this case straight lines are to be taken which are proportional to the figures, the de- monstration would this way become much longer than Eu- clid's : But, even though his demonstration bad been right, who does not see that it could not be made use of in the 5th book i PROP. F, G, H,K. B. V. These propositions are annexed to the 5th book, because they are frequently made use of by both ancient and modern geometers : And in many cases, compound ratios cannot be brought into demonstration, without making use of them. Whoever desires to see the dodlrine of ratios delivered in this 5th book solidly defended, and the arguments brouirht against it by And. Tacquet, Alph. Borellus and others, fully refuted, may read Dr. Barrow's mathematical lectures, viz. the 7th and 8th of the year 1666. The 5th book being thus corrected, I most readily agree to Wh^t the learned Dr. Barrow says,* " That there is nothing I «in ♦Page 336. 3i8 NOTES; " in the whole body of the Elements of a more subtile inven- " tion, nothing more solidly established, and more accurately *' handled, than the do£lrine of proportionals." And there is some ground to hope, that geometers will think that this could not have been said with as good reason,- since Theon's time till the present. ^ ^ Book VI. DEF. II. and V. ofB. VI. X HE 2d definition does not seem to be Euclid's but some unskilful editor's : For there is no mention made by Euclid, nor, as far as I know, by any other geometer, of reciprocal figures: It is obscursly expressed, which made it proper to render it more distinct : It would be better to put the follow- ing definition in place of it, viz. DEF. II. Two magnitudes are said to be reciprocally proportional to two others, when one of the first is to one of the other magni- tudes, as the remaining one of the last two is to the remaining one of the first. But the 5th definition, whicb, since Theon's time has been kept in the elements, to the great detriment of learners, is now justly thrown out of them, for the reasons given in the notes on the 23d prop, of this book. PROP. I. and II. B. VI. To the first of these a corollary is added, which is often Ksed : And the enunciation of the second is made more general. PROP. Ill B. VI. A SECOND case of this, as useful as the first, is given in prop. A; viz. the case in which the exterior angle of a trian- gle is bise<5led by a straight line : The demonstration of it is . very like to that of the first case, and upon this account may, probably, have been left out, as also the enunciation, by some unskilful editor. At least, it is certain, that Pappus makes use of this case, as an elementary proposition, without a de- monstration of it, in Prop. 39, of his 7th Book of Mathema- tical Colledtions. NOTES. PROP. VII. B. VI. To this a case is added which occurs not unfrcquentljr in demonstration. PROP. VIII. B. VI. ' It seems plain that some editor has changed the demonstra- tion that Euclid gave of this proposition : For, after he has demonstrated, that the triangles are equiangular to one ano- ther, he particularly shews that their sides about the equal angles are proportionals, as if this had not been done in the demonstration of the ^th prop, of this book : This superfluous part is not found in the translation from the Arabk, and is now left out. PROP. IX. B. VI. This is demonstrated in a particular case, viz. that in which the third part of a straight line is required to be cut ofF;. which is not at all like Euclid's manner : Besides, the author of thi demonstration, from four magnitudes being proportionals, con eludes that the third, of them is the same multiple of the fourth, which the first is of the second j now, this is no where demon- strated in the 5th book, as we now have it : But the editor assumes it from the confused notion which the vulgar have of proportionals : On this account it was necessary to give general and legitimate demonstration of this proposition. e a PROP. XVIII. B. VI. The demonstration of this seems to be vitiated : For the proposition is demonstrated only in the case of quadrilateral iigures, without mentioning how it may be extended to figures ot five or more sides : Besides, from two triangles being equi- angular, it is inferred, that a side of the one is to the homolo- gous side of the other, as another side of the first is to the side homologous to it of the other, without permutation of the proportionals ; which is contrary to Euclid's manner, as is clear from the next proposition: And the same fault occurs again in the conclusion, where the sides about the equal angles are not shewn to be proportionals, by reason of again neglefl- ing permutation. On these accounts, a demonstration is given in Euclid's manner, like to that he makes use of in the 20th 3 prop. -320 NOTES. Book VI. prop, of this boolc ; and it is extended to five- sided figures, by '^^''^^'^^ which it may be seen how to extend it to figures of any num- ber of sides. PROP. XXIII. B. VI. Nothing is usually reckoned more difficult in the elements of geometry by learners, than the doctrine of compound ratio, which Theon has rendered absurd and ungeometrical, by sub- stituting the 5th definition of the 6th book in place of the right definition, which without doubt Eudoxus or Euclid gave, in its proper place, after the definition of triplicate ratio, &c. in the 5th book. Theon's definition is this ; a ratio is said to be compounded of ratios 'orx* ai TuvyoXuii myiyiMryirts zp* sxvras iroXKx'nKxaiaty^itaxt wo/^yfT; t;»« :, Which Commandine thus translates : " Quando rationum quantitates inter se multi- " plicatae aliquam efficient rationem ;" that is, when the quantities of the ratios being multiplied by one another make a certain ratio. Dr. Wallis translates the word w^jX/xornTfj ** rationem exponentes," the exponents of the ratios : And Dr. Gregory renders the last words of the definition by " illius ** facit quantitatem," makes the quantity of that r,atio : But in whatever sense tne " quantities," or " exponents of the ratios,'* and their '' multiplication," be taken, the definition will be ungeometrical and useless : For there can be no multiplication but by a number : Now the quantity or exponent of a ratio (according to Eutochius in his Comment, on Prop. 4. Book 2, of Arch, de Sph. et Cyl. and the moderns explain that term) is the number which multiplied into the consequent term of a ratio produces the antecedent, or, which is the same thing, the ■number which arises by dividing the antecedent by the conse- quent ; but there are many ratios such, that no number can arise from the division of the antecedent by the consequent ; ex. gr. the ratio of which the diameter of a square has to the side of it } and the ratio which the circumference of a circle has to its diameter, and such like. Besides, that there is not the least mention made of this definition in the writings of Euclid, Archimedes, Apollonius, or other ancients, though they fre- . quentJy make use of compound ratio : And in this 23d prop. ' of the 6th book, where compound ratio is first mentioned, there is not one word which can relate to this definition, though here, if in any place, it was necessary to be brought in ; but the right definition is expressly cited in these words: " But the " ratio of K to M is compounded of the ratio of K to L, ^ " aiid NOTES. 321 " and of the ratio of L to M." This definition therefore of Book vi. Theon is quite useless and absurd : For that Theon brought it ""'^■'^^ into the Elements can scarce be doubted ; as it is to be found in his commentary upon Ptolemy's Mf7«x»> ivirx^is, page 62. where he also gives a childish explication of it, as agreeing only to such ratios as can be expressed by numbers ; and from this place the definition and explication have been exactly co- pied and prefixed to the definitions of the 6th book, as appears from Hervagius's edition : But Zambertusand Commandine, in their Latin translations, subjoin the same to these defini- tions. Neither Campanus, nor, as it seems, the Arabic ma- nuscripts, from which he made, his translation, have this de- finition. Claviiis, in his observations upon it, rightly judges, that the definition of compound ratios might have been made after the same manner in which the definitions of duplicate and triplicate ratio are given, viz. " That as in several magni- " tudes that are continual proportionals, Euclid named the " ratio of the first to the third, the duplicate ratio of the first " to the second ; and the ratio of the first to the fourth, the *' triplicate ratio of the first to the second, that is, the ratio *' compounded of two or three interjnediate* ratios that are " equal to one another, and so on ; so, in like manner, if *' there be several magnitudes of the same kind, following one " another, which are not continual proportionals, the first is *' said to have to the last the ratio compounded of all the inter- " mediate ratios, only for this reason, that these inter- " mediate ratios are interposed betwixt the two extremes, viz. " the first and last magnitudes ; even as, in the loth definition " of the 5th book, the ratio of the first to the third was called " the duplicate ratio, merely upon account of two ratios " being interposed betwixt the extremes, that are equal to one *' another : so that there is no difference betwixt this com- ** pounding of ratios, and the duplication or triplication of ** them which are defined in the 5th book, but that in the du- ** plication, triplication, &£c. of ratios, all the interposed ratios " are equal to one another j whereas, in the compounding of " ratios, it is not necessary that the intermediate ratios should *' be equal to one another," Also Mr. Edmund Scarburgh, in his English translation of the first six books, page 238, 266, expressly affirms, that the 5:h definition of the 6th book is suppositious, and that the true definition of compound rr.tio is contained in the lOth definition of the 5th book, viz. the Y deftnition 322 NOTES. BooklVI. definition of duplicate ratio, or to be understood from it, to '""^^^"^ wit, in the same manner as Clavius has explained it in the preceding citation. Yet these, and the rest of the moderns, do notwithstanding retain this 5th def. of the 6:h book, and il- lustrate and explain it by long commentaries, when they ought rather to have taken it quite away from the Elements. For, by comparing def. 5, book 6, with prop. 5, book 8, it will clearly appear that this definition has been put into the Elements in place of the right one, which has been taken out of them : Because, in prop. 5, book 8, it is demonstrated, that the plane number of which the sides are C, D, has to the plane number of which the sides are E, Z (see Hervagius's or Gre- . gory's edition), the ratio which is compounded of the ratios of their stdes j that is, of the ratios of C to E, and D to Z ; and by de;. 5. book 6, and the explication given of it by all the commentators, the ratio which is compounded of the ratios of C to E, and D to Z, is the ratio of the product made by the multiplication of the antecedents C, D, to the product by the consequents E, Z, that is, the ratio of the plane number of which the sides are C, D, to the plane number of which the sides are E, Z. Wherefore the proposition which is the 5th def. of book 6, is the very same with the 5thprop. of book 8, and therefore it ought necessarily to be cancelled in one of these places ; because it is absurd that the same proposition should stand as a definition in one place of the Elements, and be demonstrated in another place of them. Now, there is no doubt that prop. 5, book 8, should have a place in the Elements, as the same thing is demonstrated in it concerning plane num- bers, which is demonstrated in prop. 23, book 6, of equiangu- lar parallelograms ; wherefore def. 5, book 6, ought not to b? in the Elements. And from this it is evident that this defi- nition is not Euclid's, but Theon's, or some other unskilful geometer's. But nobody, as far as I know, has hitherto shewB the true use of compound ratio, or for what purpose it has been intro^ duced into geometry; for every proposition in which com- pound ratio is made use of, may without it be both enun- ciated and derr\onstrated. Now the use of compound ratio consists wholly in this, that by means of it, circumlocutions piay be avoided, and thereby propositions may be more briefly either enunciated or demonstrated, or both may be done; for instance, if thi? 23d proposition of the sixth book were to be enunciated, without mentioning compound ratio, it might be done NOTES. 323 done as follows : If two parallelograms be equiangular, and ^'^^ ^^• if as a side of the first to a side of the second, so any assumed ^■*''"^'"*^ straight line be njade to a second straight line j and as the other side of the first to the other side of the second, so the second straight line be made to a third. The first parallelo- gram is to the second, as the first straight line to the third. And the demonstration would be exactly the same as we now have it. But the ancient geometers, when they observed this enunciation could be made shorter, by giving a name to the ratio which the first straight line has to the last, by which name the intermediate ratios might likewise be signified, of the first to the second, and of the second to the third, and so on, if there were more of them, they called this ratios of the first to the last, the ratio compounded of the ratio of the first to the second, and of the second to the third straight line; that is, in the present example, of the ratios which are the same with the ratios of the sides, and by this they expressed the proposition more briefly thus : If there be two equiangular parallelograms, they have to one another the ratio which is the same with that which is compounded of ratios that are the same with the ratios of the sides ; which is shorter than the preceding enunciation, but has precisely the same meaning. Or yet shorter thus : Equiangular parallelograms have to one another the ratio which is the same with that which is com- pounded of the ratios of their sides. And these two enuncia- tions, the first especially, agree to the demonstration which is now in the Greek. The proposition may be more briefly de- monstrated, as Candalla does, thus : Let ABCD, CEFG be two equiangular parallelograms, and complete the parallelogram CDHG i then, because there are three parallelograms, AC, CH, CF, the first AC (by the definition of compound ratio) has to the third CF, the ratio which is compounded of the ratio of the first AC to the second CH, and of the ratio of CH to the third CF ; but the parallelogram AC is to the pa- rallelogram CH, as the straight line BC to CG ; and the parallelo- gram CH is to CF, as the straight line CD is to CE j therefore the parallelogram AG has to CF the ratio which is compounded of ratios that are the same with the ratios of the sides. And to this demonstration agrees the enunciation which is at present in the text, viz. equiangular Y z parallelograms 324 N O f E S. Book VI. parallelograms have tooheanothertheratio which is compound- ^'^^^''^ ed of the ratios of the sides : for the vulgar reading, " which ** is compounded of their sides," is absurd. But, in this edi- tion, we have kept the demonstration which is in the Greek text, though not so short as Candalla's ; because the way of finding the ratio which is compounded of the ratio of the sides, that is, of finding the ratio of the parallelograms, is shewn in that, but not in Candalla's demonstration ; whereby beginners may learn, in like cases, how to find the ratio which is com- pounded of two or mc;re given ratios. P'rom what has been said, it may be observed, that in any magnitudes whatever of the same kind A, B, C, D, &c. the ratio compounded of the ratios of the first to the second, of the second to the third, and soon to the last, is only a name or expression by which the ratio which the first A has to the last D is signified, and by which at the same time the ratios of all the magnitudes A to B, B to C, Cto D, from the first to the last, to one another, whether they be the same, or be not the same, are indicated ; as in magnitudes which are continual proportionals A, B, C, D, &c. the duplicate ratio of the first to the second is only a name, or expression by which the ratio of the first A to the third C is signified, and by which, at the same time, is shewn, that there are two ratios of the magni- tudes from the first to the last, viz. of the first A to the second B, and of the second B to the third or last C, which are the same with one another ; and the triplicate ratio of the first to the second is a name or expression by which the ratio of the first A to the fourth D is signified, and by which, at the same time, is shewn, that there are three ratios of the magnitudes from the first to the last, viz. of the first A to the second B, and of B to the third C, and of C to the fourth or last D, which are all the same with one another; and so in the case of any other multiplicate ratios. And that this is the right explication of the meaning of these ratios is plain from the defi- nitions of duplicate and triplicate ratio, in which Euclid makes use of the word Xtycritiy is said to be, or is called ; which word he, no doubt, made use of also in the definition of compound ratio, which Theon, or some other, has expunged from the Elements; for the very same word is still retained in the wrong definition of compound ratio, which is now the 5th of the 6th book : But in the citation of these definitions it is sometimes retained, as in the demonstration of prop. 19, book 6, NOTES. 325 6, **'the first is said to have, "fxiiy keyirxij to the third the du- Book. vi. ** plicate ratio," &c. which is wrong translated by Comman- ^<^^^^-' dine and otherSj *' has," instead of " is said to have :" and sometimes it is left out, as in the demonstracion of prop. 33, of the iith book, in which we find, "the first has, l^uy to the " third the triplicate ratio;" but without doubt l^u., "has," in this place, signifies t'ps same as ij^t^, ^eysrxiy is said to have ; so likewise in Prop. 23, B. 6, we iiud this citation, " but the *' ratio of K to M is compounded, c-vyxnTxi of the ratio of K to ' ** L, and the ratio L to M," which is a shorter way of expressing th^^ same thing, which, according to the definition, ought to have been expressed by ovyx«/o-3»< ^eytrxij is said to be compounded. From these remarks, together with the propositions sub- joined to the 5th book, all that is found concerning compound ratio, either in the ancient or modern geometers, may be un- derstood and explained. PROP. XXIV. B. VI. It seems that some unskilfiil editor has made up this demon stration as we now have it, out ef two others ; one of which- may be made from the 2d prop, and the other from the 4th of this book. For after he has, from the 2d of this book, and composition and permutation, demonstrated, that th jsidesabout the angle common to the two parallelograms are proportionals, he might have immediately concluded, that the sides about the otherequal anoles were proportionals, viz. from Prop. 34, B. i, and Prop. 7, B. 5. This he does not, but proceeds to shew, that the triangles and parallelograms are equiangular : and in a tedious way, by help of Prep. 4. of this book, and the 22d of book 5, deduces the same conclusion: FrOm which it is plain, that this ill-composed demonstration is not Euclid's : These superfluous things are now left out, and a more simple demon- stration is given from the 4ih prop, of this book, the same which is in the translation from the Arabic, by help of the 2d prop, and composition ; but in this the author neglects permu- tation, and does not shew the parallelograms to be equiangular, as is proper to do for the sake of beginners. PROP. XXV. B. VI. It is very evident that the demonsiiation which Euclid had given of this proposition has been vitiated by some unskilful hand : For, after this editor had demonstrated, that, " as the •* rectilineal figure ABC is to the rectilineal KGH, so is the Y 3 " parallelogram 3i6 NOTES. Book VI. " parallelogram BE to the parallelogram EF ;" nothing more ^*^v*-' should have been added but this, ^* and the rectilineal figure ' M ABC is equal to the parallelogram BE ; therefore the recti- ** lineal KGH is equal to the parallelogram EF," viz. from prop. 14, book 5. But betw^ixt these two sentences he has inserted this ; " wherefore, by permutation, as the rectilineal " figurcABCtothe parallelogram BE, so is the rectilineal KGH ** to the parallelogram EF ; by which, it is plain, he thought it was not so evident to conclude, that the second of four pro- portions is equal to the fourth from the equality of the first and third, which is a thing demonstrated in the 14th prop, of B. 5, as to conclude that the third is equal to the fourth, from the equality of the first and second, which is no where demon- strated in the Elements as we now have them : But though this proposition, viz. the third of four proportionals is equal to the fourth, if the first be equal to the second, had been given in the Elements by Euclid, as very probably it was, yet he would not have made use of it in this place ; because, as was said, the conclusion could have been immediately deduced without this superfluous step by permutation : This we have shewn at the greater length, both because it affords a certain proof of the vitiation of the text of Euclid ; for the very same blunder is found twice in the Greek text of prop. 23, book ii, and twice in prop. 2, book 12, and in the 5, 1 1, 12, and 18th of that book ; in which places ©f book 12, except the last of them, it is rightly left out in the Oxford edition of Command ine's translation ; And also that geometers may beware of making use of permu- tation in the like cases : for the moderns not unfrequently com- mit this mistake, and among others Commandine himself in his commentary on prop. 5, book 3, p. 6, b. of Pappus Alexandri- nus, and in other places : The vulgar notion of proportionals has, it seems, pre-occupied many so much, that they do not sufficiently understand the true nature of them. Besides, though the rectilineal figure ABC, to which ano- ther is to be made similar, may be of any kind whatever; yet in the demonstration the Greek text has "triangle" instead of *' reftilineal figure," which error is corrected in the above- named Oxford edition. PROP. XXVII. B. VI. The second case of this has axx^r, otherwise, prefixed ta it, as if it was a different demonstration, which probably has been done by some unskilful librarian. Dr. Gregory has rightly NOTES. 327 rightly left it oat : The scheme of this second case ought to ^^^ ^'i- be marked with the same letters of the alphabet which arc in '*"*^ the scheme of the first, as is now done. PROP. XXVIII, and XXIX. B. VI. These two problems, to the first of which the 27th prop, is necessary, are the most general and useful of all in the Ele- ments, and are most frequently made use of by the ancient geometers in the solution of other problems ; and therefore arc very ignorantly left out by Tacquet and Dechales in their editiv)ns of the Elements, who pretend that thev are scarce of any use : The cases of these problems, wherein it is required to apply a rectangle which shall be equal to a given square j to a given straig:it line, either deficient or exceeding by a square j are very often made use of by geometers : And, on this account, it is thought proper, for the sake of beginners, to give their constructions as follows : I. To apply a rectangle which shall be equal to a given square, to a given straight line, deficient by a square : But the given square must net be greater than that upon the half of the given line. Let AB be the given straight line, and let the square upon the given straight line C be that to which the rectangle to be applied must be equal, and this square by the determination is not greater than tHat upon half of the straight line AB. Bisedt AB in D, and if the square upon AD be equal to the square upon C, the thing required is done : But if it be not equal to it, AD must be greater than C, according to the determination : Draw DE at right angles to AB and make it equal to C ; produce ED to F, so that EF be equal to AD or DB, and from the centre E, at the distance EF, describe a circle meeting AB in G, and upon GB describe the square GBKH, and complete the re<Stangle AGHL ; also join EG : And because AB is bisected in D, the redtangle AG, GB together with the square of DG is equal* to (the square of DB, that is, of EF or EG, that is . 5 2 Y4 to) ' ■ H K /-^ A J) /G B C 328 NOTES. Book VI. to) the squares of ED, DG : Take away the square of DG '^^^'"^'^ from each of these equals ; therefore the remaining re£langle AG, GB is equal to the square of ED, that is, of C : But the rectangle AG, GP is the redlangle AH, because GH is equal to GB } therefore h ereftangle AH is equal to the given square ' upon the straight une C. Wherefore the reftangle AH, equal to the given square upon C, has been applied to the given straight line AB, deficient by the square GK. Which was to be done. 2. To apply a reftangle which shall be equal to a given square, to a given straight line, exceeding by a square. Let AB be the given straight line, a,nd let the square upon the given straight line C be that to which the redlangle to be applied must be equal. Bisect AB in D, and draw BE at right angles to it, so that BE be equal to C ; and having joined DE, from the centre D at the distance DE describe a circle meeting AB produced in G ; upon BG describe the square BGHK, and complete the rect- angle AGHL. And because AB is bisedled in D, and produced to G, the redlangle AG, GB together with the square of DB •6. 2. is equal* to (the square of DGj or DE, that is, to) the squares ofEB,BD. From each of these ^ equals take the square of DB ; therefore the remaining redfangle AG, GB is equal to the square of BE, that is, to the square upon C. But the rectan- gle AG,GB is the redtangle AH, because GH is equal toGB. Therefore the reCtangle AH is equal to the square upon C. Wherefore the rectangle AH, equal to the given square upon C, has been applied to the given straight line AB, exceeding by the square GK. Which was to be done. 3. To apply a rectangle to a given straight line which shall be equal to a given reftangle, and be deficient by a square. l?ut the given redtangle must not be greater than the square upon the half of the given straight line. Let AB be the given straight line, and let the given reftan-, gle be that which is contained by the straight lines C, D, which is not greater than the square upon the half of AB ; it is required to apply to AB a rectangle equal to the redbngle C, D, deficient by a square. Draw i 329 Book VI. '3.3. " 6.2. N O T £ S. Draw AE, BF at right angles to AB, upon the same side of it, and make AE equal to C, and BF to D : join EF and bisedl it in G ; and from the centre G> at the distance Gii, describe a circle meeting AE again in H : Join HF, and draw GK pa- rallel to it, and GL parallel to AE, meeting AB in L. Because the angle EHF in a semicircle is equal to the right angle EAB, AB and HF are parallels, and AH and BF are parallels ; wherefore AH is equal to BF, and the reciangle EA, AH equal to the redangle EA, BF, that is, to the rediangle C, D : And because EG, GF are equal to one ano» ther, and AE, LG, BF parallels ; therefore AX- and LB are equal, also EK is equal to KH^ and the re6langle C, D, from the determination, is not greater than the square of AL, the half of AB ; wherefore the rectangle EA, AH is not greater than the square of AL, that is, of KG: Add to each the square of KE i therefore the square'' of AK is not greater than the squares of EK, KG, that is, than the square of EG ; and consequently the straight line AK or. GL is not greater than GE. Now, if GE be equal to GL, the circle EHF touches AB in L, and there- fore the square of AL is = equal to the rectangle EA, AH, that is, to the given reel- angle C, D : and that which was required is done : But if EG, GL be unequal, EG must be the greater : and therefore the circle EHF cuts the straight line AB : let it cut it in the points M, N, and upon NB describe the square NBOP, and complete the redangle ANPQ^: Because LM is equal to** 1 3 3 LN, and it has been proved that AL is equal to LB ; there- fore A Vi is equal to XB, and the re£langle AN, NB equal to the redangle NA, AM, that is, to the rectangle "^ EA, AH, «Cor.36.S. or the rectangle C, D : But the rectangle AN, NB is the rectangle AP, because PN is equal to N B : Therefore the rectangle AP is equal to the rectangle C, D ; and the rectangle AP equal to the given rectangle C, D, has been applied to the given straight line AB, deficient by the square BP. Which was to be done. 4.T^ «36. 3, 330 Book VI. '55. 3. NOTES. 4. To apply a redangle to a given straight line that shall be equal to a given redangle, exceeding by a square. Let AB be the given straight line, and the rectangle C, D the given redangle, it is required to apply a redtangle to AB equal to C, D, exceeding by a square. Draw AE, BF at right angles to AB, on the contrary sides of it, and maice AE equal to C, and BF equal to D : Join EF, and bised it in G ; and from the centre G, at the distance GE, describe a circle meeting AE again in Hj join HF, and draw GL parallel to AE ; let the circle meet AB pro- duced in M, N, and upon BN describe the square BNOP, and complete the redlangle ANPQ_; because the angle EHF in a semicir- cle is equal to the right angle EAB, A3, and HF are pa- rallels, and therefore AH and BF are equal, and the rect- angle EA, AH equal to the redlangle EA, BF, that is, to the redangle C, D : And because ML is equal to LN, and AL to LB, therefore MA is equal to BN, and the reiStan- gle AN, NB to MA, AN, that is, ^ to the redangle EA, AH, or the rectangle C, D : Therefore the redangle AN, NB, that is, AP, is equal to the redangle C, D ; and to the given straight line AB the rectangle AP has bees applied equal to the given re(5tangle C, D, exceeding by the square BP. Which was to be done. Willebrordus Snellius was the first, as far as I know, who gave these constructions of the 3d and 4th Problems in his Apolionius Batavus : And afterwards the learned Dr. Halley gave them in the Scholium of the i8th Prop, of the 8th Book of ApoUonius's Conies restored by him. The 3d Problem is otherwise enunciated thus ; To cut a given straight line AB in the point N, so as to make the rectangle AN, NB, equal to a given space: Or, which is the same thing, having given AB the sum of the sides of a red- angle, and the magnitude of it being likewise given, to find its sides. And the 4th Problem is the same with this, To find a point N in the g^ven straight line AB produced, so as to make the 3 rectangle NOTES. 331 reftangle AN, NB equal to a given space : Or, which is the ^<»k VL same thing, having given AB the difference of the sides of a ^^'"'^^^ rectangle, and the magnitude of it, to find the sides. PROP. XXXI. B. VI. In the demonstration of this, the inversion of proportionals is twice negledlcd, and is now added, that the conclusion may be legitimately made by help of the 24.th Prop, of B. 5, as Clavius had done. PROP. XXXII. B. VI. The enunciation of the preceding 26th Prop, is not general enough ; because not only two similar parallelograms that have an angle common to both, are about the same diameter ; but likewise two similar parallelograms that have vertically oppo- site angles, have their diameters in the same straight lines : But there seems to have been another, and that a diredl demon- stration of these cases, to which this 32d Proposition was needful ; And the 32d may be otherwise, and something more briefly demonstrated, as follows : PROP. XXXII, B. VI. If two trianprles which have two sides of the one, Sec. Let GAF, HFC be two triangles which have two sides AG, GF, proportional to the two sides FH, HC, viz. AG to GF, as FH to HC ; and let AG be pa- rallel to FH, and GP to HC i AF and JFC are in a straight line. Draw CK parallel to FH, and Jet it meet GF produced in K; Because AG, KC are each of them parallel to FH, they are parallel'' to one another, and therefore the B^ -^7 -^ *3ai. alternate angles AGF, FKC are equal: And AG is to GF, as (FH to HC, that is=) CK to C34 j KF; wherefore the triangles AGF, CKF are equiangular'', < g ^ * and the angle AFG equal to the angle CKF : But GFK is a straight line, therefore AF and FC are in a straight line*. - • e^ ], The 26th Prop, is demonstrated from the 32d, as follows.; If two similar and similarly placed parallelograms have an angle common to both, or vertically opposite angles ; their diameters are in the game straight line. First, «3K I. •CQr.19.5, * 32. 6. NOTES. First, let the parallelograms ABCD, AEFG have the angle BAD common to both, and be similar, and similarly placed ; ABCDG, AEFG are about the same diameter. Produce EF, GF, to H, K, and join FA,FC ; then because the parallelograms ABCD, AEFG are similar, DA is to AB, as GA to AE : Wherefore the re- mainder DG is^ to the remainder EB, as GA to AE : But DG is equal toFH, EBtoFIC, andAE to CF : Therefore as FH to HC, so is AG to GF ; and FH, HC are parallel to AG, GF i and the tri- angles AGF, FHC are joined at one angle in the point F j where- fore AF, FC are in the same" straight line^ Next, Let the parallelograms KFHC, GFEA which are similar and similarly placed, have theii angles KFH, GFE vertically opposite ; their diameters AF, FC are in the same straight line. Because AG, GF are parallel to FH, HC; and that AG is to GF, as FH to HC; therefore AF, FC are in the same straight line''. PROP. XXXni. B. VI. The words " because they are at the centre," are left out as the addition of some unskilful hand. In the Greek, as also in the Latin translation, the words a. sTOX^, " any whatever," are left out in the demonstration of both parts of the proposition, and are now added as quite necessary ; and, in the demonstration of the second part, where the triangle BGC, is proved to be equal to CGK, the illative particle a.(,x^ in the Greek text ought to be omitted.' The second part of the proposition is an addition of Theon's, as he tells in his commentary on Ptolemy's MsyaXn 2t/vTa^»f, p. 50. PROP. B. C. D. B. VI. These three propositions are added, because they are fre- quently made use of by geometers. DEF. NOTES. DEF. IX and XL B. XL 1 HE similitude of plane figures is defined from the equa- lity of their angles, and the proportionality of the sides about the equal angles ; for from the proportionality of the sides only, or only from the equality of the angles, the similitude of the figures does not follow, except in the case when the figures are triangles : The similar position of the sides which contain the figures, to one another, depending partly upon each of these : And for the same reason, those are similar solid figures which have all their solid angles equal, each to each, and are contained by the same number of similar plane figures : For there are some solid figures contained by similar plane figures, of the same number, and even of the same magnitude, that are neither similar nor equal, as shall be demonstrated after the notes on the lOth definition : Upon this account it was necessary to amend the definition of simi- lar solid figures, and so place the definition of a solid angle before it : And from this and the loth definition, it is suffi- ciently plain how much the Elements have been spoiled by unskilful editors. DEF. X. B. XL Since the meaning of the word "equal" is known and established before it comes to be used in this definition : therefore the proposition which is the loth definition of this book, is a theorem, the truth or falsehood of which ought to be demonstrated, not assumed ; so that Theon, or some other editor, has ignorantly turned a theorem, which ought to be demonstrated in this loth definition : That figures are similar, ought to be proved from the definition of similar> figures ; that they are equal, ought to be demonstrated from the axiom, "Magnitudes that wholly coincide are equal to " one another;" or from Prop. A. of Book 5, or the 9th Prop, or the 14th of the same Book, from one of which the equality of all kinds of figures must ultimately be deduced. In the preceding books, Euclid has given no definition of equal figures, and it is certain he did not give this : For what is called the first def. of the third book, is really a theorem in which these circles are said to be equal, that have the straight lines from the centres to the circumferences equal, which is plain, from the definition of a circle j and therefore has by" some 334 NOTES. 12.11. Book XI. some editor been improperly placed among the definitions. ^^''^"^^ The equality of figures ought not to be defined, but demon- strated ; Therefore, though it were true, that solid figures contained by the same number of similar and equal plane fi- gures are equal to one another, yet he would justly deserve to be blamed who would make a definition of this proposition, which ought to be demonstrated. But if this proposition be not true, must it not be confessed, that geometers have, for these thirteen hundred years, been mistaken in this elementa- ry matter? And this should teach us modesty, and to acknow- ledge how little, through the weakness of our minds, we are able to prevent mistakes, even in the principles of sciences which are justly reckoned amongst the most certain ; for that the proposition is not universally true, can be shewn by many examples: The following is sufficient: Let there be any plane redtilineal figure, as the triangle ABC, and from a point D within it draw* the straight Ime DE at right angles to the plane ABC ; in DE take DE, DF equal to one another, upon the opposite sides of the plane, and let G be any point in EF; join DA, DB, DC 5 EA, EB, EC ; FA, FB, FC ; GA, GB, GO : because the straight line EDF is at right angles to the plane ABC, it makes right an- gles with DA, DB, 'DC which it meets in that plane ; and in the triangles EDB, FDB, ED and DB are equal to FD and 1 DB, each to each, and they contain right angles ; therefore the base EB is equal^ to the base FB ; in the same manner EA is equal to FA, and EC to FC: And in the triangles EBA, FBA, EB, BA, are equal to FB, BA> and the base EA is equal to the base FA ; wherefoi'c the angle EBA is equal " to the angle FBA, and the tri- angle EBA equal ^ to the triangle FBA, and the other angles equal to the other angles ; there- fore these triangles are »4.1. *^«. 1. Z^4. 6. similar ^ : In the ssme manner the triangle EBC is s/milar f the NOTES. 335 the triatngle FBC, and the triangle EAC to FAC ; therefore ^o*"^ Xi. there are cwo solid figures, each of which is contained by six ^ triangles, one of them by three triangles, the common vertex of wnich is the point G, and their bases the straight lines AB, BC, C A, and by three other triangies, the common vertex of which is the point E, and their bases the same lines AB, BC, CA : The other solid is contained by the same three tri- angles the common vertex of which is G, and their bases AB, BC, CA : and by three other triangles of which the commoa vertex is the point F, and their bases the same straight lines AB, BC, CA : Nov^^ the three triangles GAB, GBC, GCA are common to both solids, and the t^ree others EAB, EBC, ECA^ ot the first solid, have been shewn equal and smiiar to the three others FAB, FBC, FC A of the other solid, each to each: therefore these two solids are contained by the sarre number of equal and similar planes : But that they are not equal is mani- fest, because the first of them is contained in the other : There- fore it is not universally true that solids are equal which are contained by the same number of equal and similar planes. Cor. From this it appears that two unequal solid angles may be contained by the same number of equal plane angles. For the solid angle at B, which is .contained by the four plane angles EBA, EBC, GBA, GBC is not equal to the solid angle at the same point B wnich is contained by the four plane angles FBA, FBC, GBA, GBC ; for this last contains the other : And each of them is contained hy four plane angles, which are equal to one another, each to each, or are the self same, as has been proved : And indeed there may be innu- merable solid angles all unequal to one another which are each of them contained by plane angles that are equal to one another, each to each: It is likewise manifest, that the before- mentioned solids are not similar, since their solid angles are not ail equal. And that there may be innumerable solid angles all unequal to one another, which are each of them contained by the same plane angles disposed in the same order, will be plain from the three following propositions. PROP. I. PROBLEM. Three magnitudes. A, B, C being given, to find a fourth «uch, that every three shall be greater than the remaining one. Let 336 NOTES. Book XI. Let D be the fourth : therefore D must be less than A, B, ^'^^'^'^^ C together : Of the three A, B, C, let A be that which is not less than either of the two B and C : And first, let B and C together be not less than A ; therefore B, C, D together are greater than A : and bacause A is not less than B j A, C, D together are greater than B : In the like manner A, B, D together are greater than C j Wherefore in the case in which B and C together are not less than A, any magnitude D which is less than A, B, C together, will answer the problem. But if B and C together be less than A ; then, because it is required that B, C, D together be greater than A, from each of these taking away B, C, the remaining one D must be greater than the excess of A above B and C : Take therefore any magnitude D which is less than A, B, C togeiher, but greater than the excess of A above B and C ; Then, B, C, D together are greater than A j and because A is greater than either B or C, much more will A and D, together with either of the two B, C be greater than the other : And, by the con- struction, A, B, C are together greater than D. Cor. If besides it be required, that A and B together shall not be less than C and D together ; the excess of A and B together above C must not be less than D, that is, D must not be greater than that excess. PROP. II. PROBLEM. Four magnitudes A, B, C, D being given, of which A and B together are not less than C and D together, and such that any three of them whatever' are greater than the fourth; it is required to find a fifth magnitude E such, that any two of the three A, B, E shall be greater than the third, and also that any two of the three C, D, E shall be greater than the third. Let A be not less than B, and C not less than D. First, Let the excess of C above D be not less than the excess of A above B : It is plain that a magnitude E can be taken which is less than the sum of C and D, but greater than the excess of C above D ; let it be taken ; then E is greater likewise than the excess of A above B ; wherefore E and B together are greater than A ; and A is not less than B ; there- fore A and E together are greater than B : And, by the hypo- tliesis, A and B together are not less than C and D together, and C and D together are greater than E > therefore likewise A and B are greater th:in E. - ' Bu^ NOTES 337 But let the excess of A above B be greater than the excess Book xi. of C above D : And because, by the hypothesis, the three B, '"^^^'^^ C, Dare together greater than the fourth A ; C and D toge- ther are greater than the excess of A above B : Therefore a magnitude may be taken which is less than C and D together, but greater than the excess of A above B. Let this magnitude be E : and because E is greater than the excess of A above B, B together with E is greater than A : And, as in the preced- ing case, it may be shewn that A together with E is greater than B, and that A together with B is greater than E : There- fore, in each of the cases, it has been shewn, that any two of the three A, B, E are greater than the third. And because in each of the cases E is greater than the excess of C above D, E together with D is greater than C ; and by the hypothesis, C is not less than D ; therefore E together with C is greater than D ; and, by the construction, C and D together are greater than E ; Therefore any two of the three C, D, E are greater than the third. PROP. III. THEOREM. There may be innumerable solid angles all unequal to one another, each of which is contained by the same four plane angles, placed in the same order. Take three plane angles. A, B, C, of which A is not less than either of the other two, and such, that A and B toge- ther are less than two right angles: and, by Problem i, and its corollar}', find a fourth angle D such, that any three what- ever of the angles A, B, C, D be greater than the remaining anjgle, and su«h, that A and B together be not less than C and D together : And, by Problem 2, find a fifth angle E such, that any two of the angles A, B, E, be greater than the third, and also that any two of the angles C, D, E be greater than Z the 338 Book XI. NOTES. the third : And because A and B together are less than two right angles, the double of A and B together is less than four right angles : But A and B together are greater than the angle E ; wherefore the double of A, B together is greater than the three angles A, B, E together, which three are conse- quently less than, four right angles > and every two of the same angles, A, B, E are greater than the third ; therefore, by prop. 23, II, a solid angle may be made contained by three plane angles, equal to the angles A, B, E, each to each. Let this be the angle F, contained by the three plane angles GFH, HFK, GFK, which afe equal to the angles A, B, E, each to each : And because the angles C, D togerher are not greater than the angles A, B together, therefore the ju.gle? C, D, E are not greater than the angles A. B, E : But these last three ^re less than four right angles, as has beendeuionstiated: where- fore also the angles C, 1), E are together less than four right angles, and every two of them are greatet than the third ; therefore a solid angle may be rriade, which shall be contained by three plane angles equal to the angles C, D, E, each to '!3. 11. each^ 1 And, by prop. 26, 11, at the point F, in the straight line FG, a solid angle may be made equal to that which is con- tained by the three plane angles that arc equal to the angles C, D, E : Let this be made, and let the angle GFK, which is equal to F', be one of the three ; and let KFL, GFL be the other two which are equal to the angles C, D, each to each. Thus there is a solid angle constituted at the point F, contained by the four plane angles GFH, HFK, KFL, GFL, which are equal to the angles A, B, C, D, each to each. Again, Find another angle M such, that every two of the three angles A, B. M be greater than the third, and also every two of the three C, D, M be greater than the third ; Ajvj, NOTES. 339 And, as in the preceding part, it may be demonstrated, that ^^^l^- the three A, B, M are less than four right angles, as also that the three C, D, M, are less than four right angles. Make therefore^ a solid angle / \^ \^\^ »23. n. at N contained by the three plane angles ONP, PNQ_, ONQ_, which are equal to A, B, M, each to each : And by prop. 26, II, make at the point N, in the straight line ON, a solid angle contained by three plane angles, of which one is the angle ONQ equal toM, and the other two are the angles QNR, ONR, which are equal to the angles C, D, each to each. Thus, at the point N, there is a solid angle contained by the four plane angles ON P, PNQ, QNR, ONR which are equal to the angles A, B, C, D, each to each. And that the two solid angles at the points P, N, each of which is contained by the above-named four plane angles, are not equal to one another, or that they cannot coin- cide, will be plain by considering that the angles GFK, ONQ : that is, the angles E, M, are unequal by the construction ; and therefore the straight lines GF, FK cannot coincide with ON, NQ, nor consequently can the solid angles, which therefore are unequal. And because from the four plane angles A, B, C, D, there can be found innumerable other angles that will serve the same purpose with the angles E and M : it is plain that innumerable other solid angles may be constituted which are each contained by the same four plane angles, and all of them unequal to one another. Q^ E. D. And from this it appears, that Clavius and other authors are mistaken, who assert that those solid angles are equal which are contained by the same number of plane angles that are equal to one another, each to each. Also it is plain, that the 26th prop, of Book 1 1, is by no means sufficiently demonstrat- ed, because the equality of two solid angles, whereof each is contained by three plane angles which are equal to one another, tach to each, is only assumed, and not demonstrated. Z 2 340 r > K XI. NOTES. PROP. I. B. XI. The words at the end of this, *' for a straight line cannot *' meet a straight line in more than one point," are left out, as an addition by some unskilful hand ; for this is to be demon- strated, not assumed. Mr. Thomas Simpson, in his notes at the end of the second edition of his Elements of Geometry, p. 262, after repeating the words of this note, adds " Now, cata it possibly shew any " want of skill in an editor (he means Euclid or Theon) tore- <' fer to an axiom which Euclid himselfhath laid down, Book i, ** N° 14," he means Barrow's Euclid, for it is the loth in the Greek, *' and not to have demonstrated, what no man can ** demonstrate ?" But all that in this case can follow from that axiom is, that, if two straight lines could meet each other in two points, the parts of them betwixt these points must coin- cide, and so they would have a segment betwixt these points common to both. Now, as it has not been shewn in Euclid, that they cannot have a common segment, this does not prove that they cannot meet in two points, from which their not having a common segment is deduced in the Greek edition : But, on the contrary, because they cannot have a common seg- ment, as is shewn in Cor. of iith Prop. Book i, of 4to edi- tion, it follows plainly, that they cannot meet in two points, which the remarker says no man can demonstrate. Mr. Simpson, in the same notes, p. 265, justly observes, that in the corollary of Prop. 1 1, Book i, 410 edit, the straight lines AB, BD, BC, are supposed to be all in the same plane, which cannot beassumed in ist Prop. Book 1 1. This, soon after the 4to edition was published, I observed and corrected as it is now in tbis edition : He is mistaken in thinking the loth axiom he mentions here to be Euclid's ; it is none of Euclid's, but is the loth in Dr. Barrow's edition, who had it from He- rigon's Cursus, vol. I. and in place of it the corollary of nth Prop. Book I, was added. PROP. II. B. XI. This proposition seems to have been changed and vitiated by some editor ; for all the figures defined in the ist Book of the Elements, and among them triangles, are, by the hypo- thesis, plane figures ; that is, such as are described in a plane ; wherefore the second part of the enunciation needs no demon- stration. Besides, a convex superficies may be terminated by th|:ee NOTES. 341 three straight lines meeting one another : The thing that ^°°|^5^ should have been demonstrated is, that two or three straight ""^^^^"""^ lines, that meet one another, are in one plane. And as this is not sufficiently done, the enunciation and demonstration are changed into those now put into the text. PROP. III. B. XI. In this proposition the following words near to the end of it are left out, viz. " therefore DEB, DFB are not straight lines ; " in the liice manner it may be demonstrated, that there can be *' no other straight line between the points D, B ;" Because from this, that two lines include a space, it only follows that one of them is not a straight line : And the force of the argu» raent lies in this, viz. if the common section of the planes be not a straight line, then two straight lines could include a space, which is absurd j therefore the common section is a straight line. PROP. IV. B. XI. The words " and the triangle AED to the triangle BEC* are omitted, because the whole conclusion of the 4th Prop. B. I, has been so often repeated in the preceding books, it was needless to repeat it here. PROP. V. B. XI. In this, near to the end, ttiinrt^u ought to be left out in the Greek text : And the word " plane" is rightly left out in the Oxford edition of Commandine's translation. PROP. Vll. B. XI. This proposition has been put into this book by some un» skilful editor, as is evident from this, that straight lines which are drawn from one point to another in a plane, are, in the preceding books, supposed to be in that plane : And if they were not, some demonstrations in which one straight line is supposed to meet another would not be conclusive, because these lines would not meet one another : For instance, in Prop. 30, B. I, the straight line GK would not meet EF, if GK. were not in the plane in which are the parallels AB, CD, and in which, by hypothesis, the straight line EF is : Besides, this 7th Proposition is demonstrated by the preceding 3d ; in which the very thing which is proposed to be demonstrated in the 7th is twice assumed, viz. that the straight line drawn from one point to another in a plane, is in that plane ; and the same thing IS assumed in the preceding 6thProjx in which the straight line Z 3 which 342 NOTES. Book XI. which joins the points B,- D that are in the plane to which AB and CD are at right angles, is supposed to be in that plane : And the 7th, of which another demonstration is given, is kept in the book merely to preserve the number of the pro- positions ; for it is evident, from the yth and 35th definitions of the ist book, though it had not been in the Elements. PROP. VIII. B. XL In the Greek, and in Commandine's and Dr. Gregory's translations, near to the end of this Proposition, are the fol- lowing words : " But DC is in the plane through BA, AD," instead of which, in the Oxford edition of Commandine's trans- lation is rightly put, *' but DC is in the plane through BD, DA." But all the editions have the tollowing words, viz. " because A, B, BD are in the plane through BD, DA, and " DC is in the plane in which are AB, BD," which are ma- nifestly corrupted, or have been added to the text ; for there was not the least necessity to go so far about to shew that DC is in the same plane in which are BD, DA, because it imme- diately follows, from Prop. 7, preceding, that BD, DA, are in the plane in which are the parallels AB, CD : I'herefore, instead of these words, there ought only to be, " because all <' three are in the plane in which are the parallels AB, CD.'* PROP. XV. B. XI. After the words, " and because BA is parallel to GH," the following are added, *' for each of them is parallel to DE, " and are not both in the same plane with it," as being mani- festly forgotten to be put into the text. PROP. XVI. B. XI. In this, near to the end, instead of the words, " but straight " lines which meet neither way," ought^o be read, " but " straight lines in the same plane which produced meet neither " way :" Because, though in citing this definition in Prop. 27, Book I, it was not necessary to mention the words, " in the *■*■ same plane," all the straight lines in the books preceding this being in the same plane; yet here it was quite necessary. PROP. XX. B. XL In this, near the beginning, are the words, " But if not, let " BAC be the greater :" But the anjjle BAG may happen to Ve equal to one of the other two : Wherefore this place should be I NOTES. 343 be read thus, " But if not, let the angle BAC be not lessthan BoosXl. " either of the other two, but greater than DAB." v<^<-"*^ At the end of this proposition it is said, " in the same man- *' ner it may be demonstrated," though there is no need of any demonstration j because the angle BAC being not less than either of the other two, it is evident that BAC together with one of them is greater than the other. PROP. XXII. B. XI. And likewise in this, near the beginning, it is said, " But ** if not, let the angles at B, E, H be unequal, and let the *' angle at B be greater than either of those at E, H ;" Whichi words manifestly shew this place to be vitiated, because the angle at B may be equal to one of the other two. They ou^ht therefore to be read thus, " But if not, let the angles at B, E, " H be unequal, and let the angle at B be not less than either " of the other two at E, H : Therefore the straight line AC " is not less than either of the two DF, GK." PROP. XXIII. B. XI. The demonstration of this is made something shorter, by not repeating in the third case the things which were demonstrated in the first ; and by making use of the construction which Campanus has given ; but he does not demonstrate the second and third cases : The construdlion and demonstration of the third case are made a little more simple than in the Greek text. PROP. XXIV. B. XI. The word *' similar" is added to the enunciation of this pro- position, because the planes containing the solids which are to be demonstrated to be equal to one another, in the 25th pro- position, ought to be similar and equal, that the equality of the solids may be inferred from Prop. C. of this Book; And in the Oxford edition of Commandine's translation, a corollary is added to Prop. 24, to shew that the parallelograms mentioned in this proposition are similar, that the equality of the solids in Prop. 25, may be deduced from the loth def. of Book ii. PROP. XXV and XXVI. B. XI. In the 25th Prop, solid figures which are contained by the same number of similar and equal plane figures, are supposed to be equal to one another. And it seems that Theon, or some Z 4 other 344 NOTES. Book xr. Other editor, that he might save himself the trouble of de- '^^^'^^ monstrating the solid figures mentioned in this proposition to be equal to one another, has inserted the loth def. of this Book, to serve instead of a demonstration: which was very ignorantly done. Likewise in the 26th Prop, two solid angles are supposed to be equal : If each of them be contained by three plane angles which are equal to one another, each to each. And it is strange enough, that none of the commentators on Euclid have, as far as I know, perceived, that something is wanting in the demon- strations of these two propositions. Clavius, indeed, in a note upon the nth def, of this Book, affirms, that it is evident that those solid angles are equal which are contained by the same number of plane angles, equal to one another, each to each, because they will coincide, if they be conceived to be placed within one another ; but this is said without any proof, nor is it always true, except when the solid angles are contained by three plane angles only, which are equal to one another, each to each : And in this case the proposition is the same with this, that two spherical triangles that are equilateral to one another, are also equiangular to one another, and can coincide : which ought not to be granted without a demonstration, Euclid does not assume this in the case of reftilineal triangles, but demon- strates in Prop. 8, Book i, that triangles which are equilateral to one another, are also equiangular to one another ; and from this their total equality appears by Prop. 4. Book i. AndMe- nelaus, in the 4th Prop, of his first Book of Spherics, explicitly demonstrates, that spherical triangles which are mutually equi- lateral, are also equiangular to one another ; from which it is easy to shew that they must coincide, providing they have their sides disposed in the same order and situation. To supply these defeats, it was necessary to add the three Propositions marked A, B, C to this Book. For the 25th, 26th, and 28th Propositions of it,and consequently eight others, viz. the 27th, 31st, 32d, 33d, 34th, 36th, 37th, and 40th of the same, which depend upon them, have hitherto stood upon an infirm foundation j a« also, the 8th, 12th Cor. of 17th and 18th of I2th Book, which depend upon the 9th definition. For it has been shewn in the notes on def. loth of this book, that solid figures which are contained by the same number of simi- lar and equal plane figures, as also solid angles that are con- tained by the same number of equal plaac angles, are not always equal to one another. It I NOTES. 3^5 Ic is to be observed that Tacquet, in his Euclid, defines Book XI. equal solid angles tobe such, " as being put within one another ^"^^v^*^ « do coincide ;" but this is an axiom, not a definition ; for it is true of all magnitudes whatever. He made this useless defini- tion, that by it he might demonstrate the 36th Prop, of this iSook, without th? help of the 35th of the same : Concerning which demonstration, see the note upon Prop. 36. PROP. XXVIII. B. xr. In this it ought to have been demonstrated, not assumed, that the diagonals are in one plane. Clavius has supplied this defect. PROP. XXIX. B. XI. There are three cases of this proposition; the first is, when the two parallelograms opposite to the base AB have a side common to both : the second is, when these parallelograms are separated from one another ; and the third, when there is a part of them common to both ; and to this last only, the de- monstration that has hitherto been in the Elements does agree. The first case is immediately deduced from the preceding 28th Prop, which seems for this purpose to have been premised to this 29ch, for it is necessary to none but to rt, and to the 40th of this book, as we now have it, to which last, it would, with- out doubt, have been premised, if Euclid had net made use of it in the 29th ; but some unskilful editor has taken it away from the Elements, and has mutilated Euclid's demonstratioa of the other two cases, which is now restored, and serves for both at once. » PROP. XXX. B. XL In the demonstration of this, the opposite planes of the solid CP, in the figure in this edition, that is, of the solid CO in Commandine's figure, are not proved to be parallel ; which it is proper to do for the sake of learners. PROP. XXXI. B. XI. There are two cases of this proposition ; the first is, when the insisting straight lines are at right angles to the bases j the other, when they are not ; the first case is divided again into two others, one of which is, when the bases are equiangular parallelograms ; the other, when they are not equiangular : The 346 NOTES. Book xr. The Greek editor makes no mention of the first of these two "^^^^ last cases, but has inserted the demonstration of it as a part of that of the other : And therefore should have taken notice of it ill a corollary ; but we thought it better to give these two cases separately : The demonstration also is made something shorter by following the way Euclid has made use of in Prop. 14, Book 6. Besides, in the demonstration of the case in which the insisting straight lines are not at right angles to the bases, the editor does not prove that the solids described in the con- struftion, are parallelopipeds, which it is not to be thought that Euclid neglected : also the words, " of which the insistmg *' straight lines are not in the same straight lines," have been added by some unskilful hand ; for they may be in the'same straight lines. PROP. XXXII. B. XI. The editor has forgot to order the parallelogram FH to be applied in the angle FGH equal to the angle LCG, which is necessary. Ciavius has supplied this. Also, in the const rudion, it is required to complete the solid of which the base is FH, and altitude the same with that of the solid CD : But this does not determine the solid to be completed, since there may be innumerable solids upon the same base, and of the same altitude : It ought therefore to be said, " complete the solid of which the base is FH, and one of " its insisting straight lines is FD :" The same corredtion must be made in the following Proposition 33. PROP. D. B. XI. It is very probable that 'Euclid gave this proposition a place in the Elements, since he gave the like proposition concerning equiangula: parallelograms in the 23d B. 6. PROP. XXXIV. B. XI. In thiS the words, uv xt t^to-rwo-ai HKiia-iv Inji rZv avrwv tvOttZ*. " of which the insisting straight lines are not in the same " straight lines," are thrice repeated : but these words ought either to be left out, as they are byClaviu;^, or, in place of them, ought to be put, " whether the insisting straight lines be, *' or be not in the same straight lines :" For the other case is without any reason, excluded j also the words Hv ra. v^^n, c " whic 1 NO T E S. 347 ** which the altitudes," are twice put for ;»ii l^trrxirxt, " of Book XL ** which the insisting straight lines ;'* which is a plam mistake . For the altitude is always at right angles to the base. PROP. XXXV. B. XI. The angles ABH, BEM are demonstrated to be right an- gles in a shorter way than in the Greek ; and in the same way ACH, DFM may be demonstrated to be right angles : Also the repetition ot the same demonstration, which begins with " in the same manner," is lett out, as it was probably added to the text by some editor : for the words " in like manner *' we mav demonstrate," are not inserted except when the demonstration is net given, or when it is something ditterent from the other if it be given, as in Prop. 26, of this Book. Campatms has not this repetition. We have given another demonstration of the corollary, besides the one in the original, by help of which the following 36ch Prop, may be demonstrated without the 35th. PROP. XXX VT B. XI. T ACQUfiT in his Euclid demonstrates this proposition with- out the help of the 35th ; but it is plain, that the solids men- tioned in the Greek text in the enunciation of the proposiyon as equiangular, are such that their solid angles are contained by three plane angles equal to one another, each to each; as is evident from the construction, ^ow Tacquet does not de- monstrate, but assumes these solid angles to be equal to one another ; for he supposes the solids to be already made, and does not give the construction by which they are made : But, . by the second demonstration of the preceding corollary, his demonstration is rendered legitimate likewise in the case where the solids are constructed as in the text. PROP. XXXVII. B. XL In this it is assuraeH, that the ratios which are triplicate of those ratios which are the same with one another, are likewise the same with one another ; and that those ratios are the same with one another, of which the triplicate ratios are the same with one another ; but this ought not to be granted without a demonstration ; n.K did Euclid assume the first and easiest of these two propositions, but demonstrated it in the case of duplicate ratios, in the 22nd Prop. Book 6. On this account, another demonstration is given of this Proposition like to that vhich Euclid gives in Prop. 22, Bock 6, as Clavius has done. • 17. n. NOTES. PROP. XXXVIII. B. XI. When it is required to draw a perpendicular from a point in one plane which is at right angles to another plane, unto this last plane, it is done by drawing a perpendicular from the point to the common section of the planes ; for this perpen- dicular will be perpendicular to the plane, by def. 4, of this Book : And it would be foolish in this case to do it by the nth Prop, of the same ; But Euclid% Apollonius, and other geo- in'otWr meters, when they have occasion for this problem, direct a editiohs. perpendicular to be drawn from the point to the plane, and conclude that it will fall upon the common section oftheplanes, because this is the very same thing as if they had made use of the construction above mentioned, and then concluded that th€ , straight line must be perpendicular to the plane : but is ex- pressed in fewer words : Some editor, not perceiving this, thought it was necessary to add this proposition, which can never be of any use to the i ith book, and its being near to the end among propositions with which it has no connexion, is a mark of its having been added to the text. PROP. XXXiX. B. XI. In this it is supposed, that the straight lines which bisect the sides of the opposite planes, are in one plane, which ought to have been demonstrated ; as is now done, B. XII. B»oK XII. JL HE learned Mr. Moore, Professor of Greek in the Uni- ^'^'^^"^ rersity of Glasgow, observed to me, that it plainly appears from Archimedes's Epistle to Dositheus, prefixed to his books of the Sphere and Cylinder, which epistle he has restored from ancient manuscripts, that Eudoxus was the author of the chief propositions in this 12th book. PROP. II. B. XII. At the beginning of this it is said, " if it be not so, the square *< of BD shall be to the square of FH, as the circle ABCD is ( " to some space either less than the circle EFGH, or greater " than it :" And the like is to be found near to the end of this proposition, as also in Prop. 5, 11, 12, 18, of this Book: Con- 3 cerning it P T E S. 34.9 «ej-ning which it is to be observed, that in the demonstration Book XII. of theorems, it is sufficient, in this and the like cases, that a ^-^v^^ thing made use of in the reasoning can possibly exist, provid- ing this be evident, though it cannot be exhibited or found by a geometrical constru«£tion : So, in this place, it is assumed, that there may be a fourth proportional to these three magni- tudes, viz. the squares of BD, FH, and the circle ABCD j because it is evident that there is some square equal to the cir- cle ABCD though it cannot be found geometrically; and to the three rectilineal figures, viz. the squares of BD, FH, and the square which is equal to the circle ABCD, there is a fourth square proportional j because to the three straight lines which are their sides, there is a fourth straight line proportional*, « 12. t and this fourth square, or a space equal to it, is the space which in this proposition is denoted by the letter S : And the like is to be under stood in the other places above cited: And it is pro- bable that this has been shewn by Euclid, but left out bysome editor ; for the lemma which some unskilful hand has ^yided to this proposition explains nothing of it. PROP. III. B. XII. In the Greek text and the translations, it is said, <* and because the two straight lines BA, AC which meet one ano- ther," &c. here the angles BAC, KHL, are demonstrated to be equal to one another, by loth Prop. B. 11, which had been done before : Because the triangle EAG was proved to ' be similar to the triangle KHL : This repetition is left out, and the triangles BaC, KHL are proved to be similar in a shorter way by Prop. 21. B. 6. PROP. IV. B. XII. A FEW things in this are more fully explained, than in the Greek text. PROP. V. B. XIL Ik this, near to the end, are the words, is t (Av§o<r^tr thi^lh, as was before shewn ;" and the same are found again in the \ end of Prop. 18, of this Book; but the demonstration -referred io, except it be in the useless lemma annexed to the 2d Prop. i« no where in these Elements, and has bctn perhaps left out ►f some editor, who has forgot to cancel those words al»o. N O T E S. PROP.Vr. B.Xil. A SHORTER demonstration is given of this; and that which is in the Greek text may be made shorter by a step than it is : For the author of it makes use of the 22d Prop, of B, 5, twice : Whereas once would have served his purpose ; because that proposition extends to any number of magnitudes which are proportionals taken two and two, as well as to three which are proportional to other three, COR. PROP. VIII. B. XII. The demonstration of this is imperfe£b, because it is nqt shewn, that the triangular pyraniids into which those upon multangular bases are divided, are similar to one another, as ought necessarily to have been done, and is done in the like case in Prop. 12 of this Book : The full demonstration of the corollary is as follows : Upon the polygonal bases ABCDE, FGHKL, let there be similar and similarly situated pyramids which have the points M, N for their vertices: The pyramid ABCDEM has to the pyramid FGHKLN the triplicate ratio of that which the side AB has to the homologous side FG. Let the polygons be divided into the triangles ABE, EBC, » 20. 6. ECD J FGL, LGH, LHK, which are similar^ each to each : "li.def.n And because the pyramids are similar, therefore"^ the triangle EAM is similar to the triangle LFN, and the triangle ABM ' 4. &. to FGN : Wherefore^ ME is to E A, as NL to LF j and a AE to EB, so is FL to LG, b.cause the triangk-s EAB, 1 ,FG are similar ; therefore, ex aequali, as ME t© EB, so is NL tc LQ; N O T E S. 351 LG : In like manner it may be shewn, that EB is to BM, as ^"'"'^ ^i^- LGto GN ; therefore, again, ex aequali, as EM to MB, so is ""'^ LN to NG : Wherefore the triangles EMB, LNG having their sides proportionals, are"^ equiangular, and similar to one '^ 5. 6. another : Therefore the pyramids which have the triangles EAB, LFG for their bases,'and the points M, N for their ver- tices, are similar^ to one another, for their solid angles are*=, oii.def.a equal, and the solids themselves are contained by the same num- ' B- li- ber of similar planes: In the same manner the pyramid EBCM may be shewn to be similar to the pyramid LGHN, and the pyramid ECDM to LHKN : And because the pyramids EABM, LFGN are similar, and have triangular bases, the py- ramid EABMhas*^ to LFGN the triplicate ratioof that which 'a. 12. EB has to the homologous side LG. And, in the same man- ner, the pyramid EBCM has to the pyramid LGHN the triplicate ratio of that which EB has to LG : Therefore as the pyramid EABM is to the pyramid LFGN, so is the pyramid EBCM to the pyramid LGHN : In like manner, as the pyra^ mid EBCM is to LGHN, so is the pyramid ECDM to the pyramid LHKN : And as one of the antecedents is to one of the consequents, so are all the antecedents to all the conse- quents : Therefore as the pvramid EABM to the pyramid LFGN, so is the whole pyramid ABCDEM to the whole pyramid FGHKLN : And the pyramid EABM has to the pyramid LFGN the triplicate ratio of that which AB has to EG ; therefore the whole pyramid has to the whole pyramid the triplicate ratio of that which AB has to tke homologous side FG. Q. E. D. PROP. XI and XIL B. XIL The order of the letters of the alphabet is not observed in these two propositions, according to Euclid's manner, and is now restored : By which means, the first part of prop: 12, may be demonstrated in the same words with the first part of Prop. II.; on this account the demonstration of that first part is left out and assumed from Prop. ii. PROP. XIII. B. XIL In this proposition, the common ssflion of a plane parallel to the bases of a cylinder, with the cylinder itself, is supposed to be a circle, and it was thought proper briefly to demon- strate it -, from whence it is sufficiently manifest, that this plane divides the cylinder into two others : And the same thing is understood to be suppjied in Prop. 14. K o r fi s. PROP. XV. B. XII. ** And complete the cylinders AX, EO," both the enun- ciation and exposition of the proposition represent the cylin- ders as well as the cones, as already described : Wherefore the reading ought rather to be, " and let the cones be ALC, " ENG i and the cylinders AX, EO." The first case in the second part of the demonstration is wanting ; and something also in the second case of that part, before the repetition of the constru<Siion is mentioned j which are now added. PROP. XVII. B. XII. In the enunciation of this proposition, the Greek word* . tu TJiv (xii^ovx a^atK^OLv am^isv C70>tt5fov ty[^a,<^ai fxi xjayoj tas: tKxv- vovos tr^at^itt nxrat. t»j» iittpttniuv are thus translated by Comman- dine and others, " in majori solidum polyhedrum describerc '*'^ quod minoris sphseras superficiem noii tangat i" that is, " to ** describe in the greater sphere a solid polyhedron which shall " not meet the superficies of the lesser sphere :" Whereby they refer the words >«»T(Z t%v tTtifanta.-: to these next to them ms iyx<yc6\os a^xifixs: But they ought by no means to be thus translated; for the solid polyhedron doth not only meet the superficies of the lesser sphere, but pervades the whole of that sphere : Therefore the aforesaid words are to be referred to TO e-Ti^fov TroXvtl^ov^ and Ought thus to be translated, viz, to describe in the greater sphere a solid polyhedron whose super- ficies shall not meet the lesser sphere; as the meaning of the proposition necessarily requires. The demonytratioh of the proposition is spoiled and muti- lafed : For some easy things are very explicitly demonstrated, while others not so obvious are not sufficiently explained ; for example, when it is affirmed, that the square of KB is greater than the double of the square of BZ, in the first demonstra- tion ; and that the angle BZK is obtuse, in the second : Both which ought to have been demonstrated : Besides, in the first demonstration, it is said, "draw Kfi from the point K, perpen- ** dicular to BD;" whereas it ought to have been said, *' join ** KV," and it should have been demonstrated, that KV is perpendicular to BD: For it is evident from the figure in Her- Y^gius's and Gregory's editions, and from the words of the demon- I NOTES. 3S3 den^onstration, that the Greek editor did not perceive that the Book xil. perpendicular drawn from the point K to the straight line BD ^-^v^^ must necessarily fall upon the point V, for in the figure it is made to fall upon the point fi, a different point from V, which is likewise supposed in the demonstration. Commandine seems to have been aware of this ; for in this figure he marks one -and the same point with two letters V, D. ; and before Commandine, the learned John Dee, in the commentary he .annexes to this proposition in Henry Rillingsley's translation of the Elements, printed at London, ann. 1570, expressly takes notice of this error, and gives a demonstration suited to the construction in the Greek text, by which he shews that the perpendicular drawn from the point K to BD, must necessarily fall upon the point V. Likewise it is not demonstrated, that the quadrilateral figures SOFT, TPRY, and the triangle YRX, do not meet the les- ser sphere, as was necessary to have been done : only Clavius, as far as I know, has observed this, and demonstrated it by a lemma, which is now premised to this proposition, something altered, and more briefly demonstrated. In a corollary of this proposition, it is supposed that a solid polyhedron is described in the other sphere similar to that which is described in the sphere BCDE ; but, as the construdtion by which this may be done is not given, it was thought proper to give it, and to demonstrate, that the pyramids in it are similar to those of the same order in the solid polyhedron described in the sphere BCDE. From the preceding notes, it is sufficiently evident how much the Elements of Euclid, who was a most accurate geo- meter, have been vitiated and mutilated by ignorant editors. The opinion which the greatest part of learned men have en- tertained concerning the present Greek edition, viz. that it is very little or nothing different from the genuine work of Euclid, has without doubt deceived them, and made them less attentive and accurate in examining that edition ; whereby several errors, some of them gross enough, have escaped their notice from the age in which Theon lived to this time. Upon which account there is some ground to hope that the pains we have taken in correcting those errors, and freeing the Ele- ments as far as we could from blemishes, will not be unac- ceptable to good judges, who can discern when demonstrations are legitimate, and when they are not. A a The 354 NOTES. BooKXit. The obje£lions which, since the first edition, have been ^'"''''^'^ made against some things in the notes, especially against the doctrine of proportionals, have either been fully answered in Dr. Barrow's Le6l. Mathemat. and in these nates ; or are such, except one which has been taken notice of in the note on Prop. I. Book II, as shew that the person who made them has not sufficiently considered the things against which they are brought; so that it is not necessary to make any further answer to these objections and others like them against Euclid's definition of proportionals, of which definition Dr. Barrow justly says in page 297 of the above named book, that ** Nisi machinls impulsa validioribus iEternum persistet incon-- *' cussa." END OF THE NOTE a. E U C L I D'S DATA, IN THIS EDITION "SEVERAL ERRORS ARE CORRECTED, AND SOME PROPOSITIONS ADDED. By ROBERT SIMSON, M. D. Emeritus Professor of Mathematics in the Unitersitj of Glasgow. LONDON^ Prifitcd for F. Wisgbave, in the Strand, Snccefsor to Mr. Nottrsc. 1806. I PREFACE. EUCLID'S DATA is the first in order of the books written by the ancient geometers to facihtate and promote the method of resolution or analysis. In the general, a thing is said to be given which is either actually exhibited, or can be found out, that is, which is either known by hypothesis, or that can be demonstrated to be known ; and the propositions in the book of Euclid's Data shew what things can be found out or known from those that by hypo- thesis are already known ; so that in the analysis or investigation of a problem, from the things that are laid down to be known or given, by the help of these propositions other things are demonstrated to be given, and from these, other things are again shewn to be given, and so on, until that which was proposed to be found out in the problem is demon- strated to be given ; and when this is done, the problem is solved, and its composition is made and derived from the compositions of the Data which were made use of in the analysis. And thus the Data of Euclid are of the most general and ne- cessary use in the solution of problems of every kind. Euclid is reckoned to be the author of the Book of the Data, both by the ancient and modem geometers ; and there seems to be no doubt of his having written a book on this subject, but which, in the course of so many ages, has been much vi- tiated by unskilful editors in several places, both in the order of the propositions, and in the definitions and demonstrations themselves. To correct the A a 3 errors 358 PREFACE. errors which are now found in it, and bring it nearer to the accuracy with which it was, no doubt, at first written by EucUd, is the design of this edition, that so it may be rendered more useful to geometers, at least to beginners who desire to learn the investigatory method of the ancients. And for their sakes, the compositions of most of the Data are subjoined to their demonstrations, that the compositions of problems solved by help of the Data may be the more easily made. Marin us the philosopher's preface, which, in the Greek edition, is prefixed to the Data, is here left out, as being of no use to understand them. At the end of it, he says, that Euclid has not used the synthetical but the analytical method in de- livering them ; in which he is quite mistaken ; for in the analysis of a theorem, the thing to be de- monstrated is assumed in the analysis ; but in the demonstrations of the Data, the thing to be de- monstrated, which is, that something or other is given, is never once assumed in the demonstration, from which it is manifest, that every one of them is demonstrated synthetically ; though indeed, if a proposition of the Data be turned into a problem, for example, the 84th or 85th in the former edi- tions, which here are the 85th and S6th, the de- monstration of the proposition becomes the analysis of the problem. Wherein this edition differs from the Greek, and the reasons of the alterations from it, will be shewn in the notes at the end of the Data. EUCLID S DATA. DEFINITIONS. i^PACES, lines, and angles, are said to be given in magni- tude, when equals to them can be found. II. A ratio is said to be given, when a ratio of a given magnitude to a given magnitude which is the same ratio with it can be found. III. Rectilineal figures are said to be given in species, which have each of their angles given, and the ratios of their sides given. Points, lines, and spaces, are said to be given in position, which have always the same situation, and which are either actually exhibited, or can be found. A. An angle is said to be given in position, which is contained by straight lines given in position. V. A circle is said to be given in magnitude, when a straight line from its centre to the circumference is given in magnitude. A circle is said to be given in position and magnitude, the centre of which is given in position, and a straight line from it to the circumference is given in magnitude. VII. Segments of circles are said to be given in magnitude, when the angles in them, and their bases, are given in magnitude. VIII. Segments of circles are said to be given in position and mag- nitude, when the angles in them are given in magnitude, and their bases are given both in position and magnitude. IX. A magnitude is said to be greater than another by a given magnitude, when this given magnitude being taken trom it, the remainder is equal to the other magnitude. Aa4 X. A. 360 EUCLID'S See N, » 1. dcf. dat. 2. SeeN. • 1. def. 11. X. A magnitude is said to be less than another by a given magni- tude, when this given magnitude being added to it, the whole is equal to the other magnitude. ' PROPOSITION I. HE ratios of given magnitudes to one another IS A B C I) Let A, B be two given magnitudes, the ratio of A to B is given. Because A is a given magnitude, there may * be found one equal to it •, let this be C : And because B is given, one equal to it may be found j let it be D j and since A is equal to C, and B to.D : therefore'' A is to B, as C to D ; and consequently the ratio of A to B is given, because the ratio of the given magnitudes C, D, which is the same with it, has been found. PROP. II. / If a given magnitude has a given ratio to another magnitude, ''and if unto the two magnitudes by " which tlie given ratio is exhibited, and the given " magnitude, a fourth proportional can be found ;" the other magnitude is given. Let the given magnitude A have a given ratio to the mag- nitude B ; if a fourth proportional can be found to the three magnitudes above named, f> is given in magnitude. Because A is given, a magnitude may be found equal to it^-, let this be C: And be- cause the ratio of A to B is given, a ratio which is the same with it may be found j let this be the ratio of the given magnitude E to the given magnitude F : Unto the magni- tudes E, F, C, find a fourth proportional D, which, by the hypothesis, can be done. Wherefore, because A is to B, as E to F j and as £ to F4 so is C to D 1 A is'' to B, as C to * Tbe figures in the margin shew the number pf the propoikioni ifi Uit other edi- tiOMt . AB C D E DATA. 361 D. But A is equal to C ; therefore" B is equal to D. The ' ^*- ^• magnitude B is therefore given*, because a magnitude D equal * !• ^^• to it has been found. The limitation within the inverted commas is not in the Greek text, but is now necessarily added ; and the same must be understood in all the propositions of the book which depend upon this second proposition, where it is not expressly mentioned. See the note upon it. PROP. III. 3. 1 F any given magnitudes be added together, their sum shall be given. Let any given magnitudes AB, BC be added together, their sum AC is given. Because AB is given, a magnitude equal to it may' be fouad j » 1. d^f. let this be DE : And because BC is given, one equal to it may be found ; j\. [y Q let this be EF : Wherefore, because AB is equal to DE, and BC equal 1) jr p to EF ; the whole AC is equal to [ " the whole DF : AC is therefore given, because DF has been found which is equal to it. PROP. IV. 4, j[ F a given magnitude be taken from a given mag- nitude ; the remaining magnitude shall be given. From the given magnitude AB, let the given magnitude AC be taken ; the remaining magnitude CB is given. Because AB is given, a magnitude equal to it may* bc^^j^ found ; let this be DE : And be- cause AC is given, one equal to it .Y C R may be found ; let this be DF : ~ ' ' ' Wherefore because AB is equal to ^^ ir i-. DE, and AC to DF ; the remain- ii fe E der CB is equal to the remainder FE. CB is therefore given% because FE which is equal to it has been found. * 4. dat. 362 E U C D I D ' S 12. PROP. V. Set N. X F of three magnitudes, the first together with the second be given, and also the second together M'ith the third ; either the first is equal to the third, or one of them is greater than the other by a given magnitude. Let AB, BC, CD be three magnitudes, of which AB toge- ther with BC, that is, AC, is given ; and also BG together with CD, that is, BD, is given. Either A B is equal to CD, or one of them is greater than the other by a given magnitude. Because AC, BD are each of them given, they are either equal to one another, or not equal, a U f [\ First, let them be equal, and be- ^ ■^ — cause AC is equal to BD, take away the common part BC ', therefore the remainder AB is equal to the remainder CD. But if they be unequal, let AC be greater than BD, and make CE equal to BD. Therefore CE is given, because BD is given. And the whole AC is given i therefore" AE the remain- * ' TT P r r\ der is given. And because EC is ^^ " ~ ^^ ii « «qual to BD, by taking BC from both, the remainder EB is equal to the remainder CD. And AE is given ; wherefore AB exceeds EB, that is, CD, by the given magnitude AE. i- PROP. VI. seeN. 1 F a magnitude has a given ratio to a part of it, it shall also have a given ratio to the remaining part of it. Let the magnitude AB have a given ratio to AC a part of it J it has also a given ratio to the remainder BC. Because the ratio of AB to AC is given, a ratio may be > 2. Hef. found* which is the same to it : Let this be the ratio of DE, a given magnitude to the given mag- nitude DF. And because DE, DFare A C B M. dat. gi^sn> th^ remainder FE is*" given : And because AB is to AC, as DE to ]) V £ <E. 5, DF, by conversion'^ AB is to BC, as DE toEF. Therefore the ratio of AB to BC is given, be- cause the ratio of the given magnitudes DE, EF, which is the same with it, has been found. CoR. DATA. 363 Cor. From this it follows, that the parts AC, CB have a fiven ratio to one another : Because as AB to BC, so is DE to ;F i by divisions AC is to CB, as DF to FE ; and DF, FE " n. 5. are given ; therefore* the ratio of AC to CB is given. * '• ^^'f- PROP. VII. 6. J.F two magnitudes which have a given ratio to ^e* >•'• one another, be added together; the whole magni- tude shall liave to each of them a given ratio. Let the magnitudes AB, BC, which have a given ratio to one another, be added together ; the whole AC has to each of the magnitudes, AB, BC a given ratio. Because the ratio of AB to BC is given, a ratio may be found* which is the same with it ; let this be the ratio of the »2. drf. given magnitudes DE, LF : And because DE, EF are given, the A ~ R C whole DF is given'': And because «> 3.dat, as AB to BC, so is DE to EF ; by ]) "R F * composition- AC is to CB as DF to "^ ^5- •^• FE; and, by conversion'^, AC is to AB, as DF to DE '."E.a. Wherefore because AC is to each of the magnitudes AB, BC, as DF to each of the others DE, EF ; the ratio of AC to each of the magnitudes AB, BC is given^. PROP. VIII. 7. IF a given magnitude be divided into two parts See n. which have a given ratio to one another, and if a fourth proportional can be found to the sum of the two masfnitudes bv which the given ratio is exhi- bited, one of tliem, and the given magnitude ; each of the parts is given. Let the given magnitude AB be divided into the parts AC,. CB, which have a given ratio to one another ; if a fourth pro- portional can be found to the above- a C 15 named magnitudes; AC and CB are ' each of them given. -p^ ^ Because the ratio of AC to CB is ^ A — ±2 given, the ratio of AB to BC is given * therefore a ratio . 7^ j^^ which 364 2. def. « 2. dat. * 4.dat. E U C L I D'S which is the same with it can be found'', let this be the rati© of the given magnitudes, DE, EF: And because the given magnitude ^ Q 33 AB has to BC the given ratio of DE ' "" to EF, if unto DE, EF, AB a fourth j) F F proportional can be found, this which "^ ~ is BC is given=^; and because AB is given, the other part AC is given''. ny^' •' . ' 'c In the same manner, and with the like limitation, if the dif- ference AC of two magnitudes AB, BC, which have a given ratio be given i each of the magnitudes AB, BC is given. PROP. IX. JVl AGNITUDES which have given ratios to the same magnitude, have also a given ratio to one another. Let A, C have each of them a given ratio to B : A has a given ratio to C. Because the ratio of A to B is given, a ratio which is the • '2. def. same to it may be found* ; let this be the ratio of the given magnitudes_D, E : And because the ratio of B to C is given, a ratio which is the same with it may be found* ; let this be the ratio of the given magnitudes F, G : To F, G, E find a fourth proportional H, if it can be done ; and because as A is to B, so is D to E; andasBtoC, so is (I'' to G, and so is) E to H ; ejc aequali, as A to C, so is D to H: Therefore A B C D E H the ratio of A to C is given% be- cause the ratio of the given magni- F Q. tudcs D and H, which is the same • with it has been found: But if a fourth proportional to F, G, E can- not be found, then it can only be said that the ratio of A to C is compounded of the ratios of A to B, and B to C, that is, of the given ratios of D to E, and F G. DATA. 36s PROP. X. 9. Xf two or more magnitudes have given ratios to one another, and if they have given ratios, though they \ye not the same, to some other magnitudes : these other magnitudes shall also ha\e given ratios to one another. Let two or more magnitudes A, B, C have given ratios to one another ; and let them have given ratios, though they be not the same, to some other magnitudes D, E, V : The mag- nitudes D, E, F have given ratios to one another. Because the ratio of A to B is given, and likewise the ratio of A to D i therefore the ra- tio of D to B is given*; but -^ ^ » 9. «Ut. the ratio of B to £ is given, r» -r therefore* the ratio of D to E is given: And because the q -p ratio of B to C is given, and also the ratio of B to E ; the ratio of E to C is given* : And the ratio of C to F is given ; wherefore the ratio of E to F is given : D, E, F have therefore given ratios to one another. PROP. XI. 22. JlF two magnitudes have each of them a given ratio to anotiber magnitude, both of them together shall have a given ratio to that other. Let the magnitudes AB, BC have a given ratio to the mag- nitude D, AC has a given ratio to the same D. Because AB, BC have each of them a given ratio to D, the ratio A R C of AB to BC is given*: And by •9.dat. composition, the ratio of AC to CB J) is given^ : But the ratio of BC to » 7. dat. D is given ; therefore* the ratio of AC to D is given. 366; E U € L I D'S 23. PROP. XII. SeeN. If the. whole have to the whole a given ratio, and the parts have to the parts given, but not the same ratios: Every one of them, whole or part, shall have to every one a ^reat ratio. Let the whole AB have a given ratio to the whole CD, and the parts AE, EB have given, but not the same, ratios to the parts CF, FD : Every one shall have to every one, whole or part, a given ratio. . Because the ratio of AE to CF is given, as AE to CF, so make AB to CG ; the ratio therefore of A B to CG is given : wherefore the ratio of the remainder EB to the remainder « 19. 5. FG is given, because it is the same' with the ratio of AB to CG: AndtheratioofEBto FDis ~ given, wherefore the ratio of FD *^ ^ Js to FG is given** ; and, by conver sion, the ratio of FD to DG is C F G D given*^ : And because AB has to ~ ' ^ each of the magnitudes CD, CG a given ratio, the ratio of CD to CG is given**, and therefore <^ the ratio of CD to DG is given : But the ratio of GD to DF is given, vi^herefore'' the ••cor. 6. ratio of CD to DF is given, and consequently"^ the ratio of ^*'- CF to FD is given ; but the ratio of CF to AE is given, as •iO. dat. ^Iso the ratio of FD to EB ; vv^herefore"^ the ratio of AE to f ^ ^^j EB is given ; as also the ratio of AB to each of them*^. The ratio therefore of every one to every one is given. 24. PROP. XIII. See N. J^ p ^l^g ^jj.j.|. ^^ three proportional straight lines has a given ratio to the third, the first shall also have a given ratio to the second. Let A, BjC be three proportional straight lines, that is, as A to B, so is B to C ; if A has to C a given ratio, A shall also have to B a given ratio. Because the ratio of A to C is given, a ratio which is the ♦ 9. def. same with it may be found^ let this be the ratio of the given ^13, s. straight lines D, Ej and between D and E find a** mean proportional «> 9. dit. •^ 6. dat. DATA. proportional F j therefore the rectangle contained by D and K is equal to the square of F, and the reft- angle D, E is given, because its sides D, E are given ; wherefore the square of F, and the straight line F is given : And because as A is to C, so is D to E } but as A to C, so is<^ the square of A to the square of B ; and as D to E, so is*= the square of D to the ^ J^ C *2 square of F : Therefore the square'^ of A is to the square of B, as the square of D to the square of F: As therefore"" the straight line A to the straight line B, so is the straight line D to the straight line F; therefore the ratio of A to B is given*, because the ratio of the given straight lines D, F, which is the same with it, has been found. 367 D F K 2. cor. 'JO. 6. MI. 5. « 22. 6. 9. daf. PROP. XIV A. XF a magnitude, together with a given magnitude, sceN. has a given ratio to another magnitude ; the excess of this other ma;>:nitude ahove a s^iven maonitude has a o;n-en ratio to the first masrnitude : And iC the excess of a mao-nitude above a srivcn ma«:nitude has a given ratio to another magnitude ; this other magnitude, together with a given magnitufle, has a given ratio to the first magnitude. Let the magnitude AB, together with the given magnitude BE, that is, AE, have a given ratio to the magnitude CD : the excess of CD above a given magnitude has a given ratio to AB. Because the ratio of AE to CD is given, as AE to CD, so mate BE to FD; therefore the ratio of BE to FD is given, and BE is given ; wherefore FD is given^' : And because as AE »0 3 ^ »«.<Jat, to CD, so is B¥. t® FD, the re- mainder A B is ''to the remainder C F D '19.5, 'CF, as AE to CD : But the ratio of AE to CD is given; therefore the ratio of AB to CF is given ; that is, CF the excess of CD above the given magni- tude FD has a given ratio to AB. Next, Let the excess of the magnitude AB above the given magnitude BE, that is, let AE havea given ratio to the mag- nitude, 368 EUCLID'S nitude CD ; CD together with a given magniiud^ has a given ratio to AB. Because the ratio of AE to CD is given, as AE to CD, so makeBEtoFDjthereforetheratioof -r? -n BE to FD is given, and BE is given ^ J^i S •2.dat. wherefore FD is given": And because as AE to CD, so is BE to FD, AB is -r, n T ' ^^- ^' to CF, as-^ AE to CD : But the ratio ^ "M— ^ of AE to CD is given, therefore the ratio of A B to CF is given : that is, CF which is equal to CD, together with the given magnitude DF, has a given ratio to AB. B. PROP. XV. SeeN. J p r^ magnitude, together with that to which ano- ther magnitude has a given ratio, be given ; the sum of this other, and that to which the first magnitude has a given ratio, is given. Let AB, CD be two magnitudes of 'which AB together with BE to which CD has a given ratio, iji given j CD is given to- gether with that magnitude to which AB has a given ratio. Because the ratio of CU to BE is given, as BF to CD, so make AE to FD ; therefore the ratio of AE to FD is given, » 2.dat. and AE is given, wherefore* FD is given : And because as BE to A Jj K •'Cor.i9.5cD,so isAE toFD: AB is" to FC, as BE to CD: And the ratio y CT> of BE to CD is piven, wherefore ' "~~ the ratio of AB to FC is given : And FD is given, that is, CD together with FC to which AB has a given ratio is given. 10. PROP. XVI. sceN. J[p t]-,g excess of a' magnitude above a given mag- nitude has a given ratio to another magnitude ; the excess of both together above a given magnitude shall have to that other a given ratio : And if the excess of two magnitudes together above a given magnitude, has to one of tliem a given ratio ; either the excess of the other above a given magnitude has to that one a given ratio ; or tiie other is given together with the magnitude to wliich tliat one has a given ratio. /Ct DATA. 369 Let the ejtcess of the magnitude AB above a given magni- tude, have a given ratio to the magnitude BC ; the excess of AC, both of them together, above the given niagnitude, has a given ratio to BC. / Let AD be the given magnitude, the excess of AB above which, viz. DB, has a given ratio to BC : And beciuse'DB has a A. T) B C given ratio to BC, the ratio of DC to CB is given% and AD is given j therefore DC, the ex- » 7. dat. cess of AC above the given magnitude AD, has a given ratio to BC. Next, Let the excess of two magnitudes AB, BC together, above a given magnitude, have to one of them BC a given ratio j ei- A D B !E C ther the excess of the other of them AB above the given magnitude shall have to BC a given ra- tio J or AB is given, together with the magnitude to which £C has a given ratio. Let AD be the given magnitude, and first let it be less than AB ; and because DC the excess of AC above AD has a given ratio to BC, DB has'' a given ratio to BC ; that is, » «or. 6. DB the excess of AB above the given magnitude AD has a ^^- given ratio to BC. But let the given magnitude be greater than AB, and make AE equal to it ; and because EC, the excess of AC above AE has to BC a given ratio, BC has' a given ratio to BE ; and ' 6. dat. because AE Is given, AB together with BE to which BC has a given ratio is given. PROP. XVIL H. J.F the excess of a magni' ade above a given mag- SeeN. iiitude has a given ratio to another magnitude ; the excess of the same first magnitude above a given magnitude, shall have a given ratio to both the magnitudes together. And if the excess of either of two magnitudes above a given magnitude has a given ratio to both magnitudes together ; the ex- cess of the same above a given magnitude shall I have a given ratio to the other. Let the excess of the magnitude AB above a given magni- tude have a given ratio to the magnitude BC ; the excess of AB above a given magnitude has a given ratio to AC. B b Let 370 EUCLID'S Let AD be the given magnitutle ; and because DB, the ex- cess of AB above AD, has a given ratio to BC ; the ratio of * 7- dat. DC to DB is given* : Make the ratio of AD to DE the same with this ratio ; therefore the ra- . T T\ T> n tio of AD to DE is given ; and ^ - V Y ^ ^ ^ •> 2. dat. AD is given, wherefore'' DE and the remainder AE are given: ' 12. 5. And because as DC to DB, so is AD to DE, AC is-^ to EB, as DC to DB i and the ratio of DC to DB is given j where- fore the ratio of AC to EB is given : And- because the ratio of EB to AC is given, and that AE is given, therefore EB the excess^of AB above the given magnitude AE, has a given ratiato AC. Next, Let the excess of AB above a given magnitude have a ■ given ratio to AB and BC together, that is, to AC ; the ex- cess of AB above a given magnitude has a given ratio to BC. Let AE be the given ma:gnitude j and because EB the excess of AB above AE has to AC a given ratio, as AC to EB so make AD to DE ; therefore the ratio of AD to DE is d g_ dat. given, as also<^ the ratio of AD to AE : And AE is given, wherefore^ AD is given : And because, as the whole AC, to e 19 5^ the whole EB, so is AD to DE, the remainder DC is*" to the remainder DB, as AC to EB ; and the ratio of AC to EB is f Cor. 6. given ; wherefore the ratio of DC to DB is given, as also*^ the tiat, ratio of DB to BC : And AD is given ; therefore DB, the excess of AB above a given magnitude AD, has a given ratio toBC. u. PROP. xvin. 1 » i . dat. F to each of two raagnitudes, which have a given ratio to one anotlier, a given magnitude be added ; the wholes shall either have a given ratio to one another, or the excess of one of them above a given magnitude shall have a given ratio to the other. Let the two magnitudes AB, CD ha,ve a Sfiven ratio to one another, and to AB let the given magnitude BE be added, and the given magnitude DF to CD; The wholes AE, CF either have a given ratio to one another, or the excess ot one of them above a given magnitude has a given ratio to the other*. Because BE, DF are each of them given, their ratio is given, and DATA. 371 and if this ratio be the same with the A "R V ratio g; AB to CD, the ratio of AE '^ "^ ~ to CF, which is the samek with the #> 7^ -tr* " 12. 5. given ratio o( AB to CD, shall be ^^-^ £- given. But if the ratio of BE to DF be not the same with the ratio of AB to CD, either it is greater than the ratio of AB to CD, or, by inversion, the ratio of DF to BE is greater than the ratio of CD to AB : First, let the ratio of BE to DF be greater A !B C- TT^ than the ratio of AB to CD j and as AB to CD, so make BG to DF ; Q J) therefore the ratio of BG to DF is given; and DF is given, therefore"^ BG is given: And '2.dat, because BE has a greater rati > to DF than ( AB to CD, that is, than) BG to DF, be is greater"^ trian BG : And because as * '0-5. AB to CD, so is BG to DF ; therefore AG is' to CF, as AB to CD : But the ratio of AB to CD is giv°n, .vhercfore the ratio of AG to CF is given ; and because Bti^ BG are each of them given, GE is given : Therefore AC', the excess of AE above a given magnitude GE, has a given ratio to CF. The other case is demonstrated in the same manner, PROP. XIX. 15. JlF from each of two magnitudes, which have a given latio to one another, a ariven maarnitude be taken, the remainders shall either have a given ratio to one another, or tlie excess of one of them above a given magnitude, shall have a given ratio to the other. Let the maixnitudes AB, CD have a criven ratio to one an- other, and from AB let the given magnitude AE be taken, and from CD the given magnitudes CF ; The remainders EB, FD shall either have a given ratio to one another, or the excess of one of them above a given magnitude shall have a ^ iji S given ratio to the other. Because AE, CF are each of C Y T> them given, their ratio is - - _'iven*i and if this ratio be the same with the ratio of AB to » 1 ,1,^, Bb2 CD, ■ 37* EUCLID'S CD, the ratio of the remainder £B to the remainder FD, » 19. 5. which is the same** with the given ratio of AB to CD, shall be given. But if the ratio of AB to CD be not the same with the ratio of AE to CF, either it is greater than the ratio of AE to CF, or, by inversion, the ratio ©f CD to AB is greater than the ratio of CF to AE : First, let the ratio of AB to CD be greater than the ratio of AE to CF, and as AB to CD, so make AG to CF : therefore the ratio of AG to CF is given, and A, E G ^ ' 2. dat. Q£ is given, wherefore' AG is given : And because the ratio of C y D AB to CD, that is, the ratio of AG to CF, is greater than the ratio of AE to CF ; AG is • l», 5. greater** than AE ; and AG, AE are given, therefore the remainder EG is given ; and as AB to CD, so is AG to CF, and so is** the remainder GB to the remainder FD j and the ratio of AB to CD is given : W herefore the ratio of GB to FD is given ; therefore GB, the excess of EB above a given magnitude EG, has a given ratio to FD. In the same man> ner the other case is demonstrated. 16. PROP. XX. XF to one of two magnitudes which have a given ratio to one another, a given magnitude be added, and from the other a given magnitude be taken ; the excess of the sum above a given magnitude shall have a given ratio to the remainder. Let the two magnitudes AB, CD have a given ratio to one another, and to AB let the given magnitude E A be added, and from CD let the given magnitude CF be taken j the excess of the sum EB above a given magnitude has a given ratio'to the remainder FD. Because the ratio of AB to CD is given, make as AB to CD, so AG to CF : Therefore the ratio of AG to CF is given, and CF is given, wherefore* AG is given: and EA is given, there- K A G P> fore the whole EG is given: And ^ because as AB to CD, so is AG q y jy ' to CF, and so is"* the remainder ' GB to the remainder FD ; the ratio of GB to FD is given. And EG is given, therefore GB, the excess of the sum EB I abo\ DATA. 373 above the given magnitude EG, has a given ratio to the remainder FD. PROP. XXI. c. XF two magnitudes have a given ratio to one ano- SeeN. ther, if a given magnitude be added to one of them, and the other be taken from a given magnitude ; the sum, together with the magnitude to which the remainder has a given ratio, is given : And the remainder is given together with the magnitude to which the sum has a sciven ratio. o* Let the two magnitudes AB, CD have a given ratio to one another ; and to AB let the given magnitude BE be added, and let CD be taken from the given magnitude FD : The sum AE is given together with the magnitude to which the remainder FC has a given ratio. Because the ratio of AB to CD is given, make as AB to CD, so GB to FD : Therefore the ratio of GB to FD is given, and FD is given, wherefore GB is given*; and BE is given, the G A R E • 3.dat. whole GE is therefore given, and because as AB to CD, so is GB F C D to FD, and so is" GA to FC ; the ' - 1?- 5. ratio of GA to FC is given : And AE together with GA is given, because GE is given ; therefore the sum AE, together with GA, to which the remainder FC has a given ratio, is given. The second part is manifest from Prop. 15. PROP. XXII. D. If two magnitudes have a given ratio to one ano- see n. ther, if from one of them a given magnitude be taken, and the other be taken from a given mag- nitude ; each of the remainders is given together with the magnitude to which the other remainder has a given ratio. Let the two magnitudes AB, CD have a given ratio to one another, and from AB let the given magnitude AE be taken, B b 3 and 374 E U C L I D'S and let CD be taken from the given magnitude CF : The re- nlainder EB is given together vvith the magnitude to which the other remainder DF has a given ratio. Because the ratio of AB to CD is -given, make as AB to CD, so AG to CF: The ratio of AG to CF is therefore given, » 2. dat. and CF is given, vi^herefore* AG is given ; and AE is given, and .A ^ ]B Q: therefore the remainder EG is given : And because as AB to Q J) ]p b 19. 5. CD, so is AG to CF : And so is" '^ the remainder BG to the remainder DF ; the ratio of BGto DF is given : And EB together vi'ith BG is given, because EG is given: Therefore the remainder EB, together with BG, to which DF the other remainder has a given ratio, is given. The second part is plain from this and Prop. 15, ' 20, See N. PROP. XXIII. If from two given magnitudes there be taken mag- nitudes which have a given ratio to one another, the remainders shall either have a given ratio to one anotlier, or the excess of one of them above a given magTiitude shall have a given ratio to the- other.' Let AB, CD be two given magnitudes, and from them let the magnitudes AE, CF, which have a given ratio to one another, be taken ; the remainders EB, FD either have a given ratio to one another, or the excess of one of them above a given magnitude has a given ratio to the other. Because AB, CD are each of them given, the ratio of AB to A , Iji jjt CD is given : And if this ratio be the same with the ratio of AE Q T Q to CF, then the remainder EB ' » 19. 5. has' the same given ratio to the remainder FD. But if the ratio of AB to CD be not the same with the ratio of AE to CF, it is either greater than it, or, by inversion, the ratio of CD to AB is greater than the ratio of CF to AE : First, let the ratio of AB to CD be greater than the ratio of AE to CP" ; and as AE to CF, so make AG to CD : there- fore the ratio of AG to CD is given, because the ratio of •> 2. dat. AE to CF is given j and CD is given, wherefore'* AG is given i' DATA. given ; and because the ratio of AB to CD is greater than the ratio of (AE to CF, that is. 37S A E GB « 10, 5. • C D than the ratio of) AG to CD; AB is greater"^ than AG: And AB, AG are given; therefore the remainder BG ' is given : And because as AE to CF, so is AG to CD, and so is* EG to FD ; the ratio of EG to FD is given : And GB ' '*• ^• is given ; therefore EG, the excess of EB above a given mag- nitude GB, has a given ratio to FD. The other case is shewn in the same way. PROP. XXIV. J 3, 1 F there be three magnitudes, the first of which has a given ratio to the second, and the excess of the second above a given magnitude has a given ratio to tlie third ; the excess of the first above a given mag- nitude shall also have a given ratio to the third. Let AB, CD, £, be the three magnitudes of which AB has a given ratio to CD ; and the excess of CD above a given magnitude has a given ratio to E : The excess of AB above a given magnitude has a given ratio to E. Let CF be the given magnitude, the excess of CD above which, viz. FD, has a given ratio to K : And because the ratio of AB to CD is given, as AB to CD, so make AG to CF ; therefore the ratio of AG to CF ' is given : And CF is given, wherefore* AG is given : And because as A B to CD, so is AG p|_ C to CF, and so is" GB to FD ; the ratio of GB ^^ to FD is given. And the ratio of FD to E is given, wherefore "^ the ratio of GB to E is given, and AG is given ; therefore GB the excess of AB above a given magnitude AG ]g J) has a given ratio to E. CoR. I. And if the first has a given ratio to the second, and the excess of the first above a given magnitude has a given ratio to the third ; the excess of the second above a given magnitude shall have a given ratio to the third. For, if the second be called the first, and the first the second, this corol- lary will be the same with the proposition. Bb4 CoR. SeeN. M9. 5. ^ 9. dat. 376 EUCLID'S Cor. 2. Also, if the first has a given ratio to the second, and the excess of a third above a given magnitude has also a given ratio to the second, the same excess shall have a given ratio to the first j as is evident from the 9th dat. 17. PROP. XXV. I F there be three magnitudes, the excess of the first whereof above a given magnitude has a given ratio to the second j and the excess of a third above a given magnitude has a given ratio to the same se- cond ; The first shall either have a given ratio to the third, or the excess of one of them above a given magnitude shall have a given ratio to the other. Let AB, C, DE be three magnitudes, and let the excesses of each of the two AB, DE above given magnitudes have given ratios to C j AB, DE either have a given ratio to one another, or the excess of one of them above a given magni- tude has a given ratio to the other. Let FB the excess of AB above the given magnitude AF have a given ratio to C ; and let GE the ex- cess of DE above the given magnitude DG a have a given ratio to C ; and because FB,GE have each of them a given ratio to C, they jp •§, dat. have a given ratio* to one another. ButtoFB, GE the given magnitudes AF, DG are » 18. dat. added ; therefore'' the whole magnitudes AB, DE have either a given ratio to one another, or the excess of one of them above a given B magnitude has a given ratio to the other. 18- PROP, XXVL C D G _l F there be three magnitudes, tlie excesses of one of which above given magnitudes have given ratios to the other two magnitudes ; these two shall either have a given ratio to one another, or the excess of one of them above a given magnitude shall have a given ratio to the other, Let DATA. 377 A C a K L • E B D F Let AB, CD, EF be three magnitudes, and let GD the excess of one of them CD above the given magnitude CG have a given ratio to AB ; and also let KD the excess of the same CD above the given magnitude CK have a given ratio to EF, either AB has a given ratio to EF, or the excess of one of them above a given magnitude has a given ratio to the other. Because GD has a given ratio to AB, as GD to AB, so make CG to HA ; therefore the ratio of CG to HA is given ; and CG is given, wherefore* HA is given : And because as • 2. dat. GD to AB, so is CG to HA, and so is" CD to HB ; the ratio k it. s. of CD to HB is given: Also because KD has a given ratio to EF, as KD to EF, so make CK to LE i ttj therefore the ratio of CK to LE is given ; and CK is given, wherefore LE* is given : And because as KD to EF, so is CK to LE, and so" is CD to LF ; the ratio of CD to LF is given : But the ratio of CD to HB is given, wherefore*^ the ratio of HB to LF is given : ""^ *^ '9. dat. and from HB, LF the given magnitudes HA LE being taken, the remainders aB, EF shall either have a given ratio to one another, or the excess of one of them above a given magnitude has a given ratio to the other**. * I9. dat. Another Demonstration. Let AB, C, DE be three magnitudes, and let the excesses of one of them C above given magnitudes have given ratios to AB and DE ; either AB, DE have a given ratio to one another, or the excess of one of them above a given magnitude has a given ratio to the other. Because the excess of C above a given magnitude Jias a given ratio to AB ; therefore * AB together with a given mag-* i*. *t. nitude has a given ratio to C : Let this given ip magnitude be AF, wherefore FB has a given ratio toC : Also because the excess of C above a given magnitude has a given ratio to DE ; therefore * DE together with a given magni- tude has a given ratio to C : Let this given magnitude be DG, wherefore GE has a given ratio to C : And FB has a given ratio to C, therefore" the ratio ' 9. «lat, of FB to GE is given : And from FB, GE the given magni- tudes AF, DG being taken, the remainders AB, DE either have a given ratio to one another, or the excess of one of them above a given magnitude has a given ratio to the other*". ' 19. dat. B G D Cl E 378 E U C L I D'S 1?. - PROP. XXVII. -1 F there be three magnitudes, the excess of tlie first of which above a given magnitude has a given ratio to the secopd ; and the excess of the second above a given magnitude has also a given ratio to the third : The excess of the first above a given maw-nitude shall have a eiven ratio to the third. » 2. dat. 19.3. 9. dat. Let AB, CD, E be three magnitudes, the excess of the first of which AB above the given magnitude AG, viz. GB, has a given ratio to CD ; and FD the excess of CD above the given magnitude CF, has a given ratio to E : the excess of AB above a given magnitude has a given ratio to E. Because the ratio of GB to CD is given, as GB to CD, so make GH to CF ; therefore the ratio of GH to (^F is given ; and CF is given, wherefore=^ GH is given ; and AG is given, wherefore the whole AH is given : And because as GB to CD, so is GH to CF, and so is ^ the re- mainder HB to the remainder FD ; the ratio of HB to FD is given : And the ratio of FD to E is given, wherefore "^ the ratio of HB to E is given : And AH is given j therefore HB the excess of AB above a given magnitude AH ^as a giveii ratio to E. A G C H - F - B D E " 24. dut. *' Otherwise, Let AB, C, D be three magnitudes, the excess EB of the first of which AB above the given magnitude AE has a given ratio to C, and the excess of C above a given magnitude has a given ratio to D ; The ex- ^ cess of AB above a given magnitude has a given ratio to D. E Because EB has a given ratio toC, and the excess ofC above a given magnitude has a y, given ratio to D; therefore'' the excess of EB above a given magnitude has a given ratio to D : Let this given magnitude be EF; therefore g FB the excess of EB above EF has a given ratio toD: And AF is given, because AE, EF c D are D A T A. 379 are given : Therefore FB the excess of AB above a given magnitude AF has a given ratio to D." PROP. XXVIII. ^ 25. I F two lines given in position cut one another, the See n. point or points in which they cut one another are given. Let two lines AB, CD, given in position, cut one another in the point E j the point E is given. Because the lines AB, CD are given in position, they have always the same situation*, and therefore the point, or points, in which they cut one another have always the same situation : And be- cause the lines AB, CD can be found % the point, or points, in which they cut one another, are liicewise found ; and therefore are given in position*. PROP. XXIX. 2d. IF the extremities of a straight line be given In position ; the straight line is given in position and magnitude. Because the extremities of the straight line are given, they can be found * : Let these be the points, A, B, between which a 4 ^^f^ a straight line AB can be drawn ''i "i.Posttt- this has an invariable position, because J^ J^ late. between two given points there can be drawn but one straight line : And when the straight line AB is drawn, its magnitude is at the same time exhibited, or given : Therefore the straight line AB is given in position and. magnitude. 38o' EUCLID'S 2'^. ' PROP. XXX. IF one of the extremities of a straight line given in position and magnitude be given ; the other ex- tremity shall also be given. Let the point A be given, to wit, one of the extremities of a straight line given in magnitude, and which lies in thd straight line AC given in position ; the other extremity is also given. Because the straight line is given in magnitude, one equal • 1. def. to it can be found"; let this be the straight line D : From the greater straight line AC cut off AB equal to the lesser D : Therefore the A, B C other extremity B of the straight line AB is found : And the point B has jy al ways the same situation ; because any —— — — — other point in AC, upon the same side of A, cuts off between it and the point A a greater or less straight line than AB, that '*• def. is, than D: Therefore the point B is given'': And it is plain another such point can be found in AC, produced upon the other side of the' point A. '*• PROP. XXXL J-F a straight line be drawn through a given point parallel to a straight line given in position ; that straight line is given in position. Let A be a given point, and BC a straight line given in position J the straight line drawn through A parallel to BC is given in position. • 31. 1. Through A draw* the straight line DAE parallel to BC; the straight D A E line DAE has always the same posi- tion, because no other straight line ]^ (^ can be drawn through A parallel to BC : Therefore the straight line DAE which has beep found • 4 def. is given^ in position. DATA. 381 PROP. XXXII. 29. X F a straight line be drawii to a given point in a straight line given in position, and makes a given angle with it ; that straight line is given in position. Let AB be a straight line given in position, and C a given point in it, the straight line drawn to C, which makes a given angle with CB, is given in position. Because the angle is given, one equal to it can be found* j let this be the angle at D, at the given point C, in the given straight A. line AB, make '' the angle ECB equal to the angle at D : There- fore the straight line EC has al- ways the same situation, because any other straight line FC, drawn to the point C, makes with CB a greater or less angle than the angle ECB, or the angle at D : Therefore the straight line EC, which has been found, is given in position. It is to be observed, that there are two straight lines EC, GC upon one side of AB that make equal angles with it, and which make equal angles with it when produced to the other side. PROP. XXXIII. 30, JLF a straight hne be dra^vn from a given point to a straight line given in position, and makes a given angle with it, that straight line is given in position. From the given point A let the straight line AD be drawn to the straight line BC given in position, and make with it a given angle ADC : AD is given in _ position. IL =A, F Through the point A draw" the \ 'Sj.t, straightlineEAF parallel to BC J and \ because through the given point A the ^ t\ tt- straightlineEAFisdrawn parallel to BC, which is given in position, EAF is therefore given in position'' : And because the straight line AD meets the parallels t. 31. da(. BC, 382 EUCLID'S. '29. 1. BC, EF, the angle is EAD equal<= to the angle ADC j anH ADC is given, wherefore also the angle EAD is given : Therefore, because the stjaight line DA is drawn to the given point A in the straight line EF given in position, and makes with It a given angle EAD, AD is given^ in position. " 32. dat. SceN. PROP. XXXIV. XF from a given point to a straight line given in po.sition, a straight line be drawn which is given in magnitude ; the same is also given in position. Let A be a given point, and BC a straight line given in position, a straight line given in magnitude, drawn from the point A toBC is given in position. Because the straight line is given in magnitude, one equal to »i. def. it can be found'^j let this be the straight line D : From the point A draw AE perpendicular to BC : J\^ and because AE is the shortest of all the straightlines which can be drawn from the point A to BC, the straight line D, to ^L which one equal is to be drawn from the B ll, C point A to BC, cannot be less than AE. J) If therefore D be equal to AE, AE is the straight line given in magnitude drawn from the given point. A to BC : And it *33. dat. is evident that AE is given in position'', because it is drawn from the given point A to BC, which is given in position, and makes with BC the given angle AEC. But if the straight line D be not equal to AE, it must be greater than it : Produce AE, and make AF equal to D ; and from the centre A, at the distance AF, describe the circle GFH, and join AG, AH : because the circle GFH is given ■^ 6 def. in position^ and the straight line BC is also given in position j therefore their intersedion G <*28. dat. is given''; and the point A is given J wherefore AG is given in '29.dat. positions that is, the straight line AG given in magnitude, (for it is equal to D) and drawn from the given point A to the straight line BC given in posi- tion, is also given in position : And in like manner AH is ^ given in position : Therefore in this case there are two straight lines I DATA. 383 lines AG, AH of the same given magnitude which can be draw" from a given peint A to a straight line BC given in position* PROP. XXXV. 52. If a straight line be drawn between two parallel straight lines given in position, and makes given jnojles with them, the straisrht line is siven in mas:- mtude. Let the straight line EF be drawn between the parallels AB, CD, which are given in position, and malce the given an- gles BEF, EFD : EF is given in magnitude. In CD take the given point G, and through G draw^ GH »3i. i. parallel to EF : And because CD meets the parallels GH, EF, the angle EFD is equal'' to the angle "2?, i. HGD : And EFD is a given angle ; A E H R wherefore the angle HGD is given : and because HG is drawn to the given point G, in the straight line CD, given in position, and makes a given angle C F G 13 HGD; the straight line HG is given in position'^ : And AB is given in position: therefore the* 32: dat. point H is given"^ ; and the point G is also given, wherefore « 28. dat. GH is given in magnitude^ : And EF is equal to it, therefore e 09 ,ij,t. EF is given in magnitude. PROP. XXXVI. 05. xF a straight hne given in magnitude be drawn SeeN. between two parallel straight lines given in posi- tion, it shall make given angles with the parallels. Let the straight line EF given in magnitude be drawn be- tween the parallel straight lines AB, CD, which are given in position : The angles \ JT j-^^ J> AEF, EFC^shall be given. ''~ Because EF is given in magnitude, a straight line equal to it can be found^: • »i.drf Let this be G: In AB take a given point p P~K. — T) H, and from it draw^ HK perpendicu- t.if lar to CD : Therefore the straight line G, ^ that A E H R /\ / ^ C F^ ^OM^ t^ND 384 EUCLID'S that is, EF cannot be less than HK : And if G be equal to HK, EF also is equal to it ; wherefore EF is at right angles to CD J for if it be not, EF would be greater thai HK, which is absurd. Therefore the angle EFD is a right, and conse- quently a given angle. But if the straight line G be not equal to HK, it must be greater than it : Produce HK, and take HL, equal to G ; and from the centre H, at the distance HL, describe the circle • 6- "Jef- MLN, and join HM, HN : And because the circle' MLN, and the straight line CD, are given in position, the points M, < 28. dat. N are '^ given : And the point H is given, wherefore the straight lines HM, HN, •29.dat. are given in position*': And CD is given in position ; therefore the angles HMN, HNM, are given in posi- ^ 'A.def. tion^: Of the straight lines HM, HN, let HN be that which is not parallel to EF, for EF cannot be parallel to both of them ; and draw EO parallel to • 34. 1. HN : EO therefore is equale to HN, that is, to G i and EF is equal to G j wherefore EO is equal to EF, and the angle " 29. 1 , EFO to the angle EOF, that is**, to the given angle HNM, and because the angle HNM, which is equal to the angle EFO, or EFD, has been found ; therefore the angle EFD, that is, " 1. def. the angle AEF, is given in magnitude'' : and consequently the angle EFC. E. PROP. XXXVH. See N. J^ f a Straight line given in magnitude be drawn from a point to a straight line given in position, in a given angle; the straight line drawn through that point parallel to the straight line given in po- sition, is given in position. Let the straight line AD given in magnitude be drawn from the point A to the straight line BC given in position, in the given angle ADC : the E A H ^ straight line EAF drawn through A pa- rallel to BC is given in position. In BC take a given point G, and draw GH . parallel to AD : And because HG is drawn B DO C to a given point G in the straight line BC give DATA. 385 given in position, in a given angle HGC, for it is equal* to* 29. 1. the given angle ADC ; HG is given in position'' : But it is " *-. dat. given also in magnitude, because it is equal to' AD, which is ^34. 1. given in magnitude ; therefore because G, one ofthe extremi- ties of the straight line GH given in position and magnitude .is given, the other extremity H is given"^ j and the straight <« 30.dat. line E AF, which is drawn through the given point H parallel to BC given in position, is therefore given^ in position. «3i. da. PROP. XXXVIII. 3-i. XF a straight line be drawn from a given point to two parallel straight hnes given in position, the ratio ofthe segments between the given point and the parallels shall be given. Let the straight line EFG be drawn from the given point E to the parallels AB, CD, the ratio of EF to EG is given. From the point E draw EHK perpendicular to CD j and because from a given point E the straight line EK is drawn to CD which is given in position, in a given angle EKC j EK is A FH B E' C KGn> C G K given in position' ; and AB, CD are given in position ; there- »33. At. fore^ the points H, K are given : And the point E is given j «> 23. dat, wherefore' EH,EK.are given in magnitude, and the ratio'^ofcgg dat them is therefore given. But as EH to EK, so is EF to EG, * 1. dat.' because AB, CD are parallels j therefore the ratio of EF to EG is given. PROP. XXXIX. 35,30. F the ratio of the segments of a straight line be- s<?eN-. tween a given point in it and two parallel straight lines, be given, if one of the parallels be given irr position, the other is also given in position. C c Froai 386 E U CL I D'S From the given point A, let the straight line AEDbe drawn to the two parallel straight lines FG, SC, and let the ratio of* the segments AE, AD be given ; if one of the parallels BC be given in position, the other FG is also given in position. From the point A draw AH perpendicular to BC, and let it meet FG in K ; and because AH is drawn from the given point A to the straight line BC given in position-, and makes a B H TJc: B }) II C » 33. dat. " 28. dat. <: 29. dat. 6 3. dat. e 30. dat. 3I.dat. given angle AHD ; AH is given=^ in •position ; and BC is lilcewise given in positii)n, therefore the point H is given'' : The point A is also given ; wherefore AH is given in magni- tude% and, because FG, BC are parallels, as AE to AD, so is AK to AH; and the ratio of AE to AD is given, wherefore the ratio of AK to AH is given ; but AH is given in magnitude, therefore* AK is given in magnitude; and it is also given in position, and the point A is given ; wherefore'' the point K is given. And because the straight line FG is drawn through the given point K parallel to BC which is given in position^ therefore'" FG is given in position. F~ir K (r Seen. 57,58. PROP. XL. XF the ratio of the segments of a straio-ht litie into which it is cut hy three parallel straight lines, he given ; if two of the parallels are given in position, the third is also given in position. -Let AB, CD, HKbe thf^e|>arallel straight lines, of which AB, CD are given in position ; and let the ratio of the seg;- meiUs DATA. 387 ments GE, GF into which the straight line GEF is' cut by the three parallels, be given ; the third parallel HK is given in position. In AB take a given point L, and draw LM perpendicular to CD, meeting HK in N ; because LM is drawn from the given point L to CD which is given in position j and makes a given angle LMD ; LM is given in position' j and CD is * i3. dit. given in position, wherefore the point Al is given" ; and the " 23.d2t.' point L is given, LM is therefore given in magnitude'^ ; and ecg.dat. because the ratio of GE to GF is given, and as GE to GP , so H G N K A E L B /" I L B H IG "Si K ( Cor. is NL to NM ; the ratio of NL to NM is given; and therefore* ) ^' "' the ratio of ML to LN is given ; but LM is given in magni- tude'', v/herefore« LM is given in magnitude : And it is also " 2 dat. given in position, and the point L is given, wherefore^ the f^o. dat. point N is given, and because the straight line HK is drawn through the given point N parallel to CD, which is given in position, therefore HK is given in positions. 1 31 da:. PROP. XLL F. J F a straight line meets three parallel straight lines s^^^"- which are givea in position, the segments into which they cut it have a given ratio. Let the parallel straight lines AB, CD, EF given in posi- tion, be cut by the straight lineGHK j the ratio of GH tg HK is given. In A B take a given point L, and J^ G ] 15 draw LM perpendicular to CD, meet- T' "T ing EF in N j therefore^ LM is given Q TT/ ^Fi T) in position J and CD, EF are given "" "" ^ » o. !: in position, wherefore the points M, N are given: And the point Lis given; ^^-^ '^ therefore*' the straight lines LMjMN' -t- K. . N F ,. ^^^. are given in magnitude ; and the ratio C c 2 388 EUCLID'S M.dat. of LMtoMN is therefore given^: But as LM to MN, s# is GH to HK J wherefore the ratio of GH to HK is given. 39. PROP. XLII. See Jf . IF each of the sides of a triangle be given in mag- nitude, the triangle is given in species. Let each of the sides of the triangle ABC be given in mag- nitude, the triangle ABC is given in species. • 52. 1. Make a triangle^ DEF, the sides of which are equal, each to each, to the given straight lines' AB, BC, CA, which can be done ; because any two of them must be greater than the third ; and let DE be equal to AB, EF, to BC, \ jy and FD to CA ; and be- cause the two sides ED, DF are equal to the two BA, AC, each to each, and the base EF equal to the base BC ; the angle "8.1. EDF is equally to the angle BAG; therefore, because the angle EDF, which is equal to the angle BAC, has been found, c \. def. ^^^ angle BAC is given'', in like manner the angles at B, C, are given. And because the sides AB, BC, CA are |iven^ * 1. dat. their ratios to one another are given'', therefore the triangle "5. dcf. ABC is given^ in species. » -iO, PROP. XLIIL J.F each of the angles of a triangle be given in magnitude, the triangle is given in species. Let each of the aiigles of the triangle ABC be given in magnitude, the triangle ABC is given in species. Takeastraightline DEgivenin position and magnitude, and at the points D, E make^* the angle EDF equal to the angle BAC, and the angle DEF equal to ABC; there- , fore the other angles EFD, BCA ■" ^' I:^ are equal, and each of the angles at the points A, B, C, is I given data: 389^ given, wherefore each of those at the points D, E, F is given : And because the straight line FD is drawn to the given point D in DE which is given in position, making the given angle EDF ; therefore DF is given in position*'. In like manner »32.dai. EF also is given in position ; wherefore the point F is given : And the points D, E are given ; therefore each of the straight lines DE, EF, JtD is given"^ in magnitude; wherefore the ^ 29. dar. triangle DEF is given in species'*: and it is similar* to the <• 42.dat. triangle ABC; which therefore is given in species. «/t.'dk. ( «• PROP. XLIV. 41. J[F one of the angles of a triangle be given, and if the sides about it have a given ratio to one another; the triangle is given in species. Let the triangle ABC have one of its angles BAG given, and let the sides BA, AC about it have a given ratio to one another ; the triangle ABC is given in species. Take a straijjht line DE given in position and magnitude, and at the point D, in the given straight line DE, make the angle EDF equal to the given angle BAG ; wherefore the angle EDF is given ; and because the straight line FD is drawn to the given point D in ED which is given in position, making the given angle EDF ; therefore * FD is given in position\ And . ^ Tv,32.d.t. because the ratio or BA to AV^ is y^ ^ given, make the ratio of ED to y^ y^ DF the same with it, and join EF; /^ 1^ y^ - 1 and because the ratio of ED to DF R C K F is given, and ED is given, therefore^ DF is given in magni- "2. da?. tude : and it is given also in position, and the point D is given, wherefore the point F is given'': and the points D, E are^^***** given, wherefore DE, EF, FD are given*^ in magnitude : * ^' t^af. and the triangle DEF is therefore given* in species ; and be- «42.d»«. cause the triangles ABC, DEF have one angle BAC equal to oneangle EDF, and the sides about these angles proportionals ; the triangles are*" similar ; but the triangle DEF is given in *^6'*' species, and therefore also the triangle ABC. Cc3 3^0 EUCLID'S See N. » 2. dat. «> 22. 5. <20. 1. OA. 5. f 22. 1. f 42. dat. 6 5.6. 42. PROP. XLV. IF tlie sides of a triangle have to one another given ratios; the triangle is given in species. Let the sides of the triangle ABC have given ratios to one another, the triangle ABC is given in species. Take a straight line D given in magnitude ; and because the ratio of AB to BC is given, make the ratio of D to E the .same with it; and D is given, therefore^ E is given. And because the ratio of BC to CA is given, to this make the ratio of E to F the same ; andE is given, and therefore^ F. And because as AB to BC, so is D to E i by composition AB and BC^together are to BC, as D anii E to E ; but as BC to CA, so is E to F ; therefore, ex asquali^, as AB and BC are to CA, so are D and E to F, and AB and BC are greater*^ than CA ; therefore D and E are greater"* than F. In the same manner any two of the three D E, F are greater than thethird. Make ^ the triangb GHK whose sides are equal to D, E, F, so that GH be equal to D, HK to E, and KG to F ; and because D, E, F, are, each of them, given, therefore GH, HK, KG are each of them given -in magnitude ; therefore the triangle GHK is given*' in spe- cies : But as AB to BC, so is (D to E, that is) GH to HK ; and as BC to CA, so is (E to F, that is) HK to KG j there- fore, ex aequali, as AB to AC, so is GH to GK. Where- fores the triangle ABC is equiangular and similar to the tri- angle GHK ; and the triangle GHK is given in species ; therefore also the triangle ABC is given in species. CoR. If a triangle is required to be made, the sides of which shall have the same rati<bs which three given straight lines D, E, F have to one another ; it is necessary that every two of them be greater than the third. D E V DATA. 391 PROP. XLVI. 43. i-F the sides of a right angled triangle about one of the acute angles have a given ratio to one ano- ther ; the triangle is given in species. Let the sides AB, BC about the acute angle ABC of the triangle ABC, which has a right angle at A, have a given ra- tio to one another j the triangle ABC is given in species. Take a straight line DE given in position and magnitude ; and because the ratio of AB to BC is given, make as AB to BC, so DE to EF J and because DE has a given ratio to EF, and DE is given, therefore* EF is given ; and because as AB '2. dat. to BC, so is DE to EF -, and AB ts less^ than BC, therefore b 19. i. DE is less' than EF. From the point D draw DG at right an- «^ a-5. ^ gles to DE, and from the cen- tre E, af the distance EF, \ describe a circle which shall ii>- JlX^"" meet DG in two points j let -n ^^ \ ^ G be either of them, and ^-* ^ join EG ; therefore the cir- cumference of the circle is given<^ in position; and the straight line DG is given ^ in^^'^f"- position, because it is drawn to the given point D in DE given "^ ''-•*'*■• in position, in a given angle ; therefore*" the point G is given, f28.dat. and the points D, E are given, wherefore DE, EG, GD areeap. dat. givens in magnitude, and the triangle DEG in species^ >» 42. d.;. And because Che triangles ABC, DEG have the angle BAG equal to the angle EDG, and the sides about the angles ABC, DEG proportionals, and each of the other angles BCA, EGD less than a right angle ; the triangle ABC is equiangular' and i -. -. similar to the triangle DEG; but DEG is given in species; therefore the triangle ABC is given in species: And in the same manner, the triangle made by drawing a straight line irom E to the other point in which the circle meets DG is given in species. C c 4 39^ E U C L 1 D'S W- PROP. XLVir. Pee N. »32. 1. b 43. «iat. ^ 32. dat. 1 2. dat. *A. 5: fe. dcf. e28. dat. h 29^ dat, i42. dat. k]8, . 1. M7. 1. If a triangle has one of its angles which is not a- right angle given, and if the sides about another angle have a given ratio to one another; the trian- gle is given in species. Let the triangle ABC have one of its angles ABC a given, but not a right angle, and let the sides B A, AC about another angle BAG have a given ratio to one another j the triangle ABC is given in species. First, Let the given ratio be the ratio of equality, that is, let the sides BA, AC, and consequently the angles ABC, ACB, be equal j and beeau.se the angle ABC is given, the angle ACB, and also the remaining* angle BAC is given ; therefore the triangle ABC is given^ in species ; and it is evident that in this case the given angle ABC must be acute. Next, Let the given ratio be the ratio of a less to a greater, that is, let the side AB adjacent to the given angle be less than the side AC : Take a straight line DE given in position and magnitude, and make the angle DEF equal to the given angle ABC; therefore EF is given'^ in position i and because the ratio of BA to AC is given, as BA to AC, so make ED to DG ; and because the ratio of ED to DG is given, and ED is given, the straight line DG is given"*, and BA is less than AC, therefore ED is less ^ than DG. From the centre D, at the distance DG, describe the circle GE, meeting EF in F, and join DFj and because the circle is given ^ in position, as also the straight line EF, the point F is given e ; and the points D, E are given ; where- fore the straight Imes, DE, EF, FD are given •* in mag- nitude, and the triangle DEF in species'. And because BA is less than AC, the angle ACB is less'' than the angle ABC, and therefore ACB is less' than a- right angle. In the same manner, DATA. manner, because-ED is less than DG or DF, the angle DFE is less than a right angle: And because the triangles ABC, DEF have the angle ABC equal to the angle DEF, and the sides about the angles BAG, EDF proportionals, and each of the other angles ACB, DFE less than a right angle ; the triangles ABC, DEF are == similar, and J3EF is given in spe- cies, wherefore the triangle ABC is also given in species. Thirdly, Let the given ratio be the ratio of a greater to a less, that is, let the said AB adjacent to the given angle be greater ihan AC ; and as in the last case, take a straight line DE given in position and magnitude, and make the angle DEF equal to the given angle ABC ; therefore EF is given^ in posi- tion : Also draw DG perpendicular to EF ; therefore if the ratio ofBA to AC be the same with the ratio of ED to the perpendicular DG, the triangles ABC, DEG are similar^, because the angles ABC, DEG are equal, and DGE is a right angle : Therefore the angle ACB is a right angle, and the triangle ABC is given in° species. But if, in this last case, the given ratio of BA to AC be not the same with the ratio of ED to DGy that is, with the ratio of BA to the perpendicular AM drawn from A to BC ; the ratio of BA to AC must be less than " the ratio of BA to AM, because AC is greater than AM. Make as B A to AC, so ED to DH ; therefore the ratio of ED to DH is less than the ratio of ( B A to AM, that is, than the ratio of) ED to DG ; and consequently DH is greater? than DG ; and because BA is greater than AC, ED is greater' than DH. From the centre D, at the distance DH, describe the circle KHF which necessa- rily meets the straight line EF in two points, because DH is greater than DG, and less than DE. Let the circle meet EF in the points F, K which are given, as was shewn in the preceding case ; and DF ; DK. being joined, the triangles DEF, DEK are given in species, as was there shewn, 393 °7,6. « 32. (Ut. * 43.dat. 394 EUCLID'S shewn; From the centre A, at the distance AC, describe a circle meeting BC again in L : And if the angle ACB be less than a right angle, ALB must be greater than a right angle ; and on the contrary. In the same manner, if the angle DFE be less than a right angle, DKE must be greater than one j and on the contrar/U Let each of the angles ACB, DFL be either less or greater than a right angle ; and because in the triangles ABC, DtF the angles ABC, DEF are equal, and the sides BA, AC, and ED, DF, about two of the other angles proportionals, the tri- ""T, 6. angle ABC is similar "" to the trianglp «DEF. Jn the same manner, the tri- angle ABL is similar to DEK, And the triangles DEF, DEK are given in species ; therefore also the triangles ABC, ABL are given in species. And from this it is evi- dent, that, in this third case, there are always two triangles of a different species, to which the things mentioned as given in the proposition* can agree. * 9. i. «> 3. 6. « 12. 5. " 47. daf. « 43, daf. -,. PROP. XLVIIL If a triangle has one angle given, and irboth the sides tooether about that an^-le liave a e'iven ratio to the remaining side ; the triangle is given in species. Let the triangle ABC have the angle BAC given, and let the sides BA, AC together about that angle have a given,ratio to BC ; the triangle ABC is givdn in species. Bisedl^ the angle BAC by the straight line AD j therefore the angle BAD is given. -And because as B A to AC, so is^ BD to DC, by permutation, as AB to BD, 8o is AC to CD ; and as BA and AC to- A gether to BC, so is'^ AB to B D. But the ratio of BA and AC together to BC is given, wherefore the ratio of AB to BD ^ is given, and the angle BAD is given ; J^ therefore ^ the triangle ABD is given in species, and the angle ABD is therefore given ; the angle BAC is also given, wherefore the triangle ABC is given in«pecies . A triangle which shall have the things that are mentioned in the proposition to be given, can be found in the following manner. D A T- ^A*^ i. 395 manner. Let EFG be the given angle, and let the ratio of H to K be the given ratio which the two sides about the angle EFG must have to the third side of the triangle ; therefore be- cause two sides of a triangle are greater than the third side, the ratio of H to K must be the ratio of a greater to a less. BisecT the angle EFG by the straight line FL, and bv the '•- ^' 47th proposition find a triangle of which EPL is one of the angles, and in which the ratio of the sides about the angle op- posite to FL is the same with the ratio of H toK: Todo which, take FE given in position and magnitude, and draw EL per- pendicular to FL : Then if the ratio of H to K be the same with the ratio of FE to EL, produce EL, and let it meet FG in P ; the triangle FEP is that which was to be found : For it has the given angle EFG ; and because this angle is bisefted by FL, the sides EF, FP together are to EP, as'' FE to EL, that i , asFItoK. But if the ratio of H to K be not the same with the ratio of FE to EL, it must be less than it, a^ was shewn in Prop. 47, and in this case there are two triangles, each of which has the given an^e EFL, and the ratio of the sides about the angle opposite to FL the same with the ratio of H to K. By Prop. 47, find these triangles EFM, ' EFN, each of which has the angle EFL for one of its angles, and the ratio of the side FE to EM or EN the same with the ratio of H to K ; and let the angle EMF be greater, and ENF less than a right angle. And because H is greater than K, EF is greater than EN, and therefore the angle EFN, that is, the angle NFG, is less^ than the angle ENF. To each of these f 13. 1. add the angles NEF, EFN ; therefore the angles NEF, EFG are less than the angles NEF, .EFN, FNE, that is, than two right ajigles j theretore the straight lines EN, FG must meec together when produced ; let them meet in O, and produce: Ei\I to G. Each of the triangles EFG, EFG has tke things mentioned to be given in the proposition : For each of them has the given angle EFG ; and because this angle is bisected by the straight line FAIN, the sides EF,FG together have to EG the third side the ratio of FE to EM, that is, of H to K. In like manner, the sides EF, FO together have to EG the ratio which H has to K. 396 E U C L I D'S w. • PROP. XLIX. If a triangle has one angle given, and if the sides about another angle, both together have a given ratio to the third side ; the triangle is given in species. Let the triansfle ABC have one angle ABC given, and let the two sides B A, AC about another angle B AC have a given ratio to BC ; the triangle ABC is giv^en in species. Suppose the angle BAC to be bisedlcd by the straight line AD ; BA and AC together are to BC, as AB to BD, as was shewn in the preceding proposition, ^ut the ratio of BA and AC together to BC is given J therefore also the ratio of A B to "44. dat. BD is given. And the angle ABD is given, wherefore* the triangle ABD is given in species ; and consequently the angle BAD; and its double the angle BAC are given : And the angle ABC is given. Therefore the triangle ABC * 43. dat. is given in species^. A triangle which shall have the things mentioned in the proposition to be given, may be thus found. Let EFG be the given angle, and the ratio -rr E ofH to K the given ratio; and by -tr ^-'^'/XllT . Prop. 44. find the triangle EFL, ^ ^^ I v'" which has the angle EFG for one of ^ / \ its angles, and the ratio of the sides p Tj O EF, FL about this angle the same with the ratio of H to K ; and make the angle LEM equal to the angle FEL. And because the ratio of H to K is the ratio which two sides of a triangle have to the third, H must be greater than K : and because EF is to FL, as H to K ; there- fore EF is greater than FL, and the angle FEL, that is, LEM, is therefore less than the angle EIE. Wherefore the angles LFE, P'EM are less than two right angles, as was shewn in the foregoing proposition, and the straight lines FL, EM muat^ meet if produced ; let them meet in G, EFG is the triangle which was to be found ; for EFG is one of its angles, and be- cause the angle FEG is bjsedled. by EL, the two sides FE, FXj together have to the third side FG the ratio of EF to FL, that is, the given ratio of H to K. D ATA. 397 PROP. L, T(i. IF from the vertex of a triangle given in species, a straight line be drawn to the base in a given angle ; it shall have a given ratio to the base. From the vertex A of the triangle ABC which is given ia species, let AD be drawn to the base BC in a given angle ADB j the ratio of AD to BC is given. Because the triangle ABC is given in species, the angle ABD is given, and the angle ADB is given, therefore the tri- angle ABD is given ' hi species ; where- / V ^V^^^ **3. dn. fore the ratio of AD to AB is given. And the ratio of AB to BC is given ; and therefore" the ratio of AD to BC is given. » 9. da;. PROP. LI. M. AXecti LINEAL figurcs given in species, are dir vided into triangles which are given in species. Let the re£lilineal figure ABCDE be given in species : ABCDE may be divided into triangles given in species. Join BE, BD ; and because ABCDE is given in species, the angle B AE is given% and the ratio jV » 3. d?f. ef BA to AE is given*; wherefore the triangle BAE is given in species", and the angle AEB is therefore given'. ( But the whole angle AED is given, and therefore the remaining angle BED is given, and the ratio of AE to EB is given, as also the ratio AE to ED ; therefore the ratio of BE to ED is given^. And the angle BED is given, where- ' fore the triangle BED is given** in species. In the same man- ner, the triangle BDC is given in species : Therefore rc^l- iineal figures which are given in species are divided into, trio" angles given in species. 398 EUCLID'S 43. PROP. Lll. If two triangles given in species be described upon the same straight hne ; they shall have a given ra- tio to one another. Let the triangles ABC, ABD given In species be described upon the same straight line AB ; the ratio of the triangle ABC to the triangle ABD is given. Through the point C, draw GE parallel to AB, and let it meet DA produced in E, and join BE. Because the triangle ABC IS given in species, the angle BAG, that is, the an*;le ACE, is given; and because the triangk ABD is given in species, the angle ?. 1. DA B, that is, the El__ ■. C angle AEC, is given. XherefoFe the trian- gle ACE is given in. species ; wherefore the. ratio of EA to AC is » 3. clef. given% and the ratio of CA to AB is given, k 9. riat. as also the ratio of BA to AD ; therefore the ratio of b E A to ■^21. 1. AD is given, and the triangle ACB is equa^ to the triangle AEB, and as the triangle AEB, or ACB, is to the triangle •1 !. P. ADB, so is ^ the straight line EA to , AD : But the ratio oi'" EA to AD is given ; therefore the ratio of the triangle ACB to the triangle .ApB is given. ■^-.;. .'m1^-'' PROfiLEM; , ,,. To firii^ (he ratio of two triangles ASCj ABI) given in spe- cies, and which-are described upon the same straight line Al>. Take a straight line FG given in position and magnitude, and because the angles of the triangles ABC, ABD are given, at the points F, G of the straight line FG, make the angles GFH, GFK"= equal to the angles BAG, BAD ; and the angles FGH, FGK equal to the angles ABC, ABD, each to each. Therefore the triangles ABC, ABD are equiangular to the tri- angles FGII, FGK, each, to each. Through the point H draw HL parallel to FG, meeting KF produced in L. And because theungles BAG, BAD are equal to the angles GFH,GFK, each to each ; therefore the angles v\GE, AEC arc equal to FHL, FLH, each to each, and the triangle AEC equiangular to the triangle FLH. Therefore as EA to AC, so is LF to FH, and as DATA. 399 as CA to AB, so HF to FG j and as BA to AD, so is GF toFK ; wherefore, ex Equali, as EA to AD, so is LF to FK. But as was shewn, the triangle ABC is to the triangle ABD, as the straight line EA to AD, that is, as LF to FK. The ratio therefore of LF to FK has been fpaud, which is the same with the ratio of the triangle ABC to thCjCri^Dgle ABD. PROP. LIII. ? If two rectilineal figures given in species. be de-sesx. scribed upon the same strais:bt line ; they shall have a given ratio to one another. Let any two rectilineal figures ABCD.E,ABFG, which are given in species, be described upon the .same st^ight iijie AB i the ratio of them to one another is given. ^ ... $ / . . Join AC, AD, AF ; each of the triangles AED^ ADC, ACB, AGF, ABF is gi ven^ in species. And b^cau£« the t^- * Ji. dat. angles ADF, ADC given in spe- :^ > cies dro described upon the same D straight line AD, the ratio of EAD F^-^-""/^ \^ to DAC is given'' j and, by com- 't"^^ / S^^ " *-• °^^* position, the ratio of EACD to \ X ^^^-^''^ j DAC is given':. And the ratio -Aj^:;;;____/t» ' "• ''''^• DAC to CAB is given"*, because 7^ \ they are described upon the same Q.L ^~"~~''~~Ay straight line AC ; therefore the . ratio of EACD to ACB is given"^; | £ K L jfN q and, bv composition, the ratio of ABCDE to ABC is given. In the same manner>ther4tioor ABFG to ABF is given. But the ratio of the triangle ABC to the triangle ABF is given ; wherefore^ because the ratio of ABCDE to ABC is given, as also the ratio of ABC to ABF, and the ratio of ABF to ABFG ; the ratio of the redilioeai ABCDE to the rectilineal ABFG is given*^. PROBLEx\L To find the ratio of two rectilineal figures given in species, and described upon the same straight line. Let ABCDE, ABFG be two rectilineal figures given in species, and described upon the same straight line AB, and join AC, AD, AF, Take a straight line HK given in posicioti and magnitude, and by the 52d dat. find the ratio of the tri- angle ADE to the triangle ADC, andmalft the ratio of HK 400 £ U C L I D'S to KL the same with it. Find also the ratio of the triangle ACD to the triangle ACB. And make the ratio of KL to LM the same. Also, find the ratio of the triangle ABC to the triangle ABF, and make the ratio of LM to MN the sartie* And, lastly, find the ratio of the triangle AFB to the triangle. AFG, and make the ratio of MN to NO the same. Then the ratio of ABCDE to ABFG is the same with the ratio of HM to MO. Because the triangle £AD is to the triangle DAC, as the straight line HK to KL ; and as the triangle DAC to CAB, so is the straight line KL to LM ; therefore by using ^ _ nir-Kr composition as often as the number Jj ^ A- X^ ^^^ Q of triangles requires, the redtilineal ABCDE is to the triangle ABC, as the straight line HM to ML, In like- manner because the triangle GAF is to FAB, as ON to NM, by composition, the re6iiiineal ABFG is to the triangle ABF as MO to NM, and by inversion, as ABF to ABFG, so is NM to MO. And the' triangle ABC is to ABF, as LM to MN. Wherefore, because as ABCDE to ABC, so is HM to ML ; and as ABC to ABF, so is LM to MN ; and as ABF to ABFG, so is MN to MO ; ex iequali, as the redilincal ABCDE to ABFG, so is the straight HM to MO. .=-0. PROP. LIV. XF two straight lines have a given patio to one ano* ther ; the similar rectilineal figures described upon them similarly, shall have a giv- en ratio to one another. Let the straight lines AB, CD, have a given ratio to one an- other, and let the similar and similarly placed reftilineal figures E, F be described upon them j the ratio of E to F is given. ToAB, CD, let G be a third proportional ; therefore as AB to CD, so is CD to G. And the ratio of AB to CD is given; wherefore the ratio of CD to G is given ; and consequently the ratio of AB to G .^ (lat. is also given*. But as A B to G, so so.cT'"^* '^ ^^^ figure E to the figure^ F. Therefore the ratio of E to F is given. DATA. PROBLEM. 401 20.6. To find the ratio of two similar rectilineal figures E, F, simi- larly described upon straight lines AB,CD which have a given ratio to one another ; Let G be a third proportional to AB,CD. Take a straight line H given in magnitude ; and because the ratio of AB to CD is given, make the ratio of H to K the same with it ; and because H is given, K is given. As H is to K, so make K to L ; then the ratio of E to F is the same with the ratio of H to L ; for AB is to CD, as H to K, wherefore CD is to G, as K to L ; and, ex aequali, as AB to G, so is H to L : But the figure E is to'' the figure F, as AB *> -^^<^ to G, that is, as H to L. PROP. LV. If two straight lines have a given ratio to one ano- ther ; the rectilineal figures given in species, described upon tliem, shall have to one another a given ratio. Let AB, CD be two straight lines which have a given ratio to one another ; the rectilineal figures E, F given in species and described upon them, have a given ratio to one another. L^pon the straight line AB, describe the figure AG similar and similarly placed to the figur* F ; and because F is given in species, AG is also given in spe- Therefore, since the fi- H- B Gr Kr C D gures E, AG which are given A in species, are described upon the same straight line AB, the ratio of E to AG is given% and because the ratio of AB to CD is given, and upon them are described the similar and similarly placed rediiineal figures AG, F, the ratio of AG to b54.;dat L a 33. dat. is given i therefore F is given'' j and the ratio of AG to E the ratio of E to F i* given'. PROBLEM. To find the ratio of two redilineal fitrures E, F griven in species ^nd described upon the straight Imes AB, CD which have a given ratio to one another. Take a straight line H given in magnitude; and because the recftilineal figures E, AG given in species, are described up- on the same straight line AB, find their ratio by the 53d dat. and make the ratio of H to K the same, K is therefore given. And because the similar rectilineal figures AG, F are described D d upon « 9. dat. 401 EUCLID'S upon the straight lines AB, CD, which have a given ratio, find their ratio by the 54th dat. and make the ratio of K to L the same ; The figure E has to F the same ratio which H has to L : For, by the eonsti uftion, as E is to AG, so is H to K ; and as AG to F, so is K to L ; Therefore, ex sequali, as E to F i so is H to L. b2. 1 PROP. LVI. >F a rectilineal figure given in specves be described upan a straight line given in ,niagnitude; the fi- gure is given in magnitude. Let the re£tilineal figure ABODE given in species, be de- scribed upon the straight line AB given in magnitude j the figure ABCDE is given in magnitude. Upon AB let the square AF be described ; therefore AF is given in species and magnitude, and because the rectilineal figures ABCDE, AF given in species are described upon the same straight line AB, the ratio of ABCDE to AF is *53. dat. given* : But the square AF is given in 2. dat, maj;;nitude, therefore^ also the figure ABCDE is given in magnitude. PROB. Tofindthemagnitudeofaredlilineal figure given in species described upon a straight line given in magnitude. Take the straight line GH equal to the given straight line AB, and by the 53ddat. find the ratio which the square AF upon AB has to the figure AtsCDE ; and make the ratio of GH to HK the same i and upon GH describe tfiesquare GL, and complete the parallelogram LHKM ; thp fipiurc ABCDE is equal to LHKM j because AF is to ABCDE, as the straight line GH to HK,that is, as the figure GL to HM ; and AF is equal to GL > therefore ABCDE is equal to HM^ PROP. LVIL IF two rectilineal figures arc given in species, and if a side of one of them has a given ratio to a side of the other ; the ratios of the remaining sides to the remaining sides shall be given. Let «:]*. 5 DATA. 403 Let AC, DF be two re<£lilineal figures given In species, and let the ratio of the side AB to the side DE be given, the ratios of the remaining sides to the remaining sides are also given. Because the ratio of AB to DE is given, as also= the ratios »3. deO of AB to BC, and of DE to EF, the ratio of BC to EF is given^ In the same manner the * lO- da»- ratios of the other sides to the other sides are given. The ratio which BC has to EF may be found thus : Take a straight line G given in magni- tude, and because the ratio of BC to BA is given, make the ratio of G to H the same ; and because the ratio of AB to DE is given, make the ratio of H to K the same ; and make the ratio of K to L the same with the given ratio of DE to EF. Since therefore as BC to BA, so is G to H ; and as BA to DE, so is H to K ; and as DE to EF so is^ K to L J ex iequali, BC is to EF, as G to L j therefore the ratio of G to L has been found, which is the same with the ratio of BC to EF. PROP. LVIII. c. 2 Cor. 20.6, J F two similar rectilineal figures have a given ratio ^*^* lO one another, tlieir homologous sides have also a triven ratio to one another. o Let the two similar r&Shlineal figures A, B have a given ratio to one another, their homologous sides have also a given ratio. Let the side CD be homologous to EF, and to CD, EF let the straight line G be a third proportional. As therefore* CD » to G, so is the figure A to B ; and the ratio of A to B is given, there- fore the ratio of CD to G is given j and CD, EF, Gare proportionals ; wherefore'' the ratio of CD to EF is given. The ratio of CD to EF may be found thus : Take a straight line H given in magnitude ; aiid because the ratio of the figure A to B is given, make the ratio of H to K the same with it : And, as the 13th dat. direds to be done, find a mean propor- D d 2 tional 1^ * 13. iiu H L IC 404 EUCLID'S tional L between H and K j the ratio of CD to EF is the same with that of H to L. Let G be a third proportional to CD, EF i therefore as CD to G, so is (A to B, and so is) H to K ; and as CD to EF, so is H to L, as is shewn in the 13th dat. 54. PROP. LIX. Se«N. If two rectilineal figures given in species have a given ratio to one another, their sides shall likewise have given ratios to one another. Let the two reftilineal figures A, B, given in species, have a given ratio to one another, their sides shall also have given ratios to one another. If the figure A be similar to B, their homologous sides shall have a given ratio to one another, by the preceding proposi- tion ; and because the figures are ;^iven in species, the sides of • 3. dif. c^^h of them have given ratios* to one another ; therefore each " 9, dati side of two of them has*" to each side of the other a given ratio. But if the figure A be not similar to B, let CD, EF be any two of their sides; and upon EF conceive the figure EG to be described similar and similarly placed to the figure A, so that CD, EF be hu- mologous sides j therefore C -T) £ N. B / •*' EG is given in species ; and — the figure B is given in spe- W "^ K « 53. dat. cjgs . vvherefore = the ratio of B to EG is given j and the -ji^ ratio of A to B is given, -y^ therefore ^ the ratio of the L figure A to EG is given ; and A is similar to EG ; therefore "58. dat. ''the ratio of the side CD to EF is given ; and consequently'' the ratios of the remaining sides to the remaining sides are given. The ratio of CD to EF may be found thus: Take a straight line H given in magnitude, and because the ratio of the figure A to B is given, make the ratio of H to K the same with it. And by the 53d dat. find the ratio of the figure B to EG, and make the ratio of K to L the same : Between H and L find a mean proportional M, the ratio of CD to EF is the same with the ratio of H to Mj because the figure A is to B as H to K j aind as B to EG, so is K to L j ex %quali, as A to EG, so is H tp DATA. H to L : And t?ne figures A, EG, are similar, and M is a mean proportional between H and L j therefore, as was shewn in the preceding proposition, CD is to EF as H to M. 405 PROP. LX. 55, K If a rectilijieal figure be given in species and mag^ nitude, the sides of it shall be given in magnitude. Let the redilineal figure A be given in species and magni- tude, its sides are given in magnitude. Take a straight line BC given in position and magnitude, and upon BC describe* the figure D similar, and similar!)' * ^8. 6. placed, to the figure A, and let EF be the side of the figure A homo- logous to BC the side of D; therefore the fi- gure D is given in spe- cies. And because upon the given straight line BC the figure D given in species is described, D is given'' in magnitude, and the figure A is given in magni- ^ tg j^( tude, therefore the ratio of A to D is given : And the figure A is similar to D ; therefore the ratio of the side EF to the homologous side BC is given'^ ; and BC is given, wherefore'* 1 2%au* EF is given : And the ratio of EF to EG is given', therefore • 3. def. EG is given. And, in the same manner, each of the other sides of the figure A can be shewn t* be given. PROBLEM. To describe a reftilineal figure A similar to a given figure D, and equal to another given figure H. It is Prop. 25, B. 6.Elem. Because each of the figures D, H is given, their ratio is gi- ven, which may be found by making^ upon the given straight fcor. 45. 1. line BC the parallelogram BK equal to D, and upon its side CK making^ the parallelogram KL equvil to H in the angle KCL equal to the angle MBC ; therefore the ratio of D to H, that is, of BK to KL, is the same with the ratio of BC to CL : And because the figures D, A are similar, and that the ratio of D to A, or H, is the same with the ratio of BC to CL; by the 58th dat. the ratio of the homologous sides BC, EF is the same with the ratio of BC to the mean proportional between BC and CL. Find EF the mean proportional j then EF is the D d 3 side 4o6 EUCLID'S side of the figure to be described, homologous to BC the side of D, and the figure itself can be described by the i8th Prop. B. 6, which, by the construction, is similar to D ; and because 6 2. cor. D is to A, as 8 BC to CL, that is, as the figure BK to KL ; and ^ 20. 6. jhat D is equal to BK, therefore A'' is equal to KL, that is, to H, 57. PROP. LXI. siceK. JLF a parallelogram given in magnitude has one of its sides and one of its angles given in magnitude, the other side also is given. Let the parallelogram ABDC given in magnitude, have the side AB and the angle BAC given in magnitude, the other side AC is given. Take a straight line EF given in position and magnitude j and because the parallelogram AD is given in magnitude, a redtilincal »i.def. figure equal to it can be found*. And a parallelogram equal to this »^<3r 45. 1 figure can be applied^ to the given straight line EF in an angle equal to the given angle BAC. Let this be the parallelogram EFHG, hav- ing the angle FEG equal to the angle BAC. And because the pa- Q J{_ rallel' g ams AD, EH are equal, and have tne angles at A and E equal ; the sides about them ar<e c 14 g. reciprocally proportional'^ ; therefore as AB to EF, so is EG to AC : And AB, EF, EG are given, therefore also AC is d j2. 6. given''. Whence the way of finding AC is manifest. A r. / / C .]) ; / H. PROP. Lxn. »e« N. If a parallelogram has a given angle, the rectangle contained by the sides about that angle has a given ratio to the parallelogram. Let the parallelogram ABCD have the given angle ABC, the re«^ngle AB, 'BC has a given ratio to the parallelogram AC. From the point A draw AE perpendi- cular to BC; because the angle ABC is given, as also the angle AEB, the triangle ••43.dat. ABE is given* in species ; therefore the ft^tio of BA to AE is given. But as BA to AE, so is*" the reAangle AB, BC, to the rectangle AE, BC, therefore the ratio of I DATA. 40; the redangle AB, BC to AE, BC that is«, to the parallel©- ^ ?5. 1. gram AC is given. And it is evident how the ratio of the re£langle to the pa- rallelogram may be found, by making the angle FGH equal to the given angle ABC, and dravvmg, from any poin" F ia one of its sides, t K perpendicular to the other GH ; for GF is to FK, as BA to A£, that is, as the reitangle AB, BC, to the parallelogram AC. Cor. And if a triangle ABC has a given angle ABC, the 66. re<Stangle AB, BC contained by the sides about that angle, shall have a given ratio to the triangle ABC. Complete the parallelogram ABCD ; therefore, by this pro- position, the rectangle AB, BC has a given ratio to the paral- lelogram AC ; and AC has a given ratio to its half the tri- angle** ABC; therefore the reitangle AB,.BC has a given 4 41 1 'ratio to the triangle ABC. , q ^^^ And the ratio of ihe rectangle to the triangle is found thus : Make the triangle FGK, as was shewn in the proposition : the ratio of GF to the half of the perpendicular f K is the same with the ratio of the redangle AB, BC to the triangle ABC. Be- cause, as was shewn, GF is to FK, as AB, BC to the paralle- logram AC ; and FK is to its half, as AC is to its half, which is the triangle ABC ; therefore, ex aequali, GF is to the half of FK, as AB, BC redangle is to the triangle ABC. PROP. LXIII. 56, X F two parallelograms be equiangular, as the side of the first to a side of the second, so is the other side of the second to the straight line to which the other side of the first has the same ratio which the first parallelogram has to the second. And conse- quently, if the ratio of the first parallelogram to the second be given, the ratio of the other side of the first to that straight line is given ; and if the ratio of the other side of the first to that straight line be given, the ratio of the first parallelogram to the se- cond is given. Let AC, DF be two equiangular parallelograms, as BC, a side of the first, is to EF, a side of the second, so is DE, the other side of the second, to the straight line to which AB, the other side of the first has the same rAtio which AChastd DF. D d 4 Produce 4o8 EUCLID'S Produce the straight line AB, and make as BC to;EF, so DE to BG, and complete the parallelo- gram BGHC ; therefore, because BC J ^ or GH, is to EF, as DE to BG, the r 7 sides aDout the equal angles BGH, DEF ^/ / are reciprocally proportional ; where- r^ 7'C a 14. 6. fore* the parallelogram BH is equal to (J^i- tr DF ; and AB is to BG, as the paral- j lelogram AC is to BH, that is, to PF; -y 7 as therefore BC is to EF, so is DE to / / BG, which is the straight line to which K^ F A.B has the same ratio that AC has to DF. ^ And if the ratio of the parallelogram AC to DF be given, then the ratio of the straight line AB to BG is given ; and if the ratio of AB to the straight line BG be given, the ratio of the parallelogram AC to DF is given. 74.73. PROP. LXIV. See N. Iy two parallclograms have unequal but given an- gles, and if as a side of the first to a side of the se- cond, so the other side of the second be made to a certain straight. hne ; if the ratio of the first paral- lelogram to the second be given, the ratio of the other side of the first to that straiglit line shall be given. And if the ratio of the other side of the first to that straight line be given, the ratio of the first parallelogram to the second shall be given. Let ABCD, EFGH be two parallelograms which have the unequal but given angles ABC, EFG; and as BC to FG, so make EF to the straight line M. If the ratio of the paral- lelogram AC to EG be given, the ratio of AB to M is given. At the point B of the straight line BC make the angle CBK equal to the angle EFG, and complete the parallelogram KBCL. And because the ratio of AC to EG is given, and that » 35. 1 . AC is equal* to the parallelogram KC, therefore the ratio of KC to EG is fr'iven ; and KC, EG are equiangular j there- "CS. dat. fore as BC to FG, so is** EF to the straight line to which KB has a given ratio, viz. the same which the parallelogram KC has to EG ; but as BC to FG, so is EF to the straight liiie M i therefore KB has a given ratio to M ; and the ratio DATA, of AB to BK is given, because the triangle ABK is given In species'^ ; therefore the ratio of AB to M is given"*. ^ q^^^** And if the ratio of AB to M be given, the ratio of the ' ** parallelogram AC to EG is given : for since the ratio of KB to BA is given, as also the ratio of AB ^ to M, the ratio of KB to M is given"* ; IV^A.^ and because the parallelograms KC, EG are etiuianguiar, as BC to FG, so is *'EF to the straight line to which KB has the same ratio which the parallelo- gram KC has to EG ; but as BC to FG, so is EF to Al ; therefore KB is to M, as the parallelogram KC is to EG ; and the ratio of KB to M is given, therefore the ratio of the pa- :j. rallelogram KC, that is, of AC to EG^ is given. Cor. And if two triangles ABC, EFG, have two equal angles, or two unequal, but given, angles ABC, EFG, and if as BC a side of the first to FG a side of the second, so the other side of the second EF be made to a straight line M; if the ratio of the triangles be g^iven, liie ratio of the other side of the first to the straight line M is given. Complete the parallelograms A BCD, EFGHj and because the ratio of the triangle ABC to the triangle EFG is given, the ratio of the parallelogram AC to EG is given', because the pa- « 15. 5. ralielograms are double^ of the triangles : and because BC is to '41. i. FG, as EF to M, the ratio of AB to M is given by the 63d dat. if the angles ABC, EFG are equal j but if they be un- equal, but given angles, the ratio of AB to M is given by this proposition. And if the ratio of AB to M be given, the ratio of the pa- rallelogram AC to EG is given by the same propositions ; and therefore the ratio of the triangle ABC to EFG is given. PROP. LXV. I F two equiano^ular parallelograms have a given 6i. ratio to one another, and if one side has to one side a given ratio ; tlie other side shall also have to the rrt' Other side a given ratio. O' Let the two equiangular parallelograms AB, CD have a given ratio to one another, and let the side EB have a given ratio to the side FD ; the other side AE has afso a given ra- tio to the other side CF. Because B F i) 410 E U C L I D'S Because the two equiangular parallelograms AB, CD have a given ratio to one another: as EB, a side of the first, is to FD, » 63. dat. a side of the second, so is^ FC, the other side of the second, to the straight line to which AE, the other side of the first, has the same given ratio which the first parallelogram AB has to the other CD. Let this straight line be EGj therefore the ratio of AE to EG is given ; and EB is to FD, as FC to \ ^ ^>^ EG, therefore the ratio of FC to EG is given, because the ratio of EB to FD is given ; and because the ratio of AE to EG, as also the ratio of FC to EG is given ; the ratio ■ jFT R T '' 9. dat. of AE to CF is given''. The ratio of A E to CF may be found thus : Take a straight line H given in magnitude ; and because the ratio of the paral- lelogram AB to CD is given, make the ratio of H to K the same with it. And because the ratio of FD to EB is given, make the ratio of K to L the same : The ratio of AE to CF is the same with the ratio of H to L. Make as EB to FD, so FC to EG, therefore, by inversion, as FD to EB, so is EG to FC ; and as AE to EG, so is'' (the parallelogram AB to CD, and so is) H to K ; but as EG to FC, so is (FD to EB, and so is) K to L i therefore, ex aequali, as AE to FC, so is H to L. «f. PROP. LXVI. J. F two parallelograms have unequal but given an- gles, and a given ratio to one another; if one side has to one side a given ratio, the other side has also a given ratio to the other side. Let the two parallelograms A BCD, EFGH which have the given unequal angles ABC, EFG have a given ratio to one another, and let the ratio of BC to FG be given ; the ratio also of AB to EF is given. At the point B of the straight line BC make the angle CBK ; equal to the given angle EFG, and complete the parallelo- gram BKLC ; and because each of the angles BAK, AKB isj , 1 dat g'^^"> ^^^ triangle ABK is given* in species j therefore thej ratio of AB to BK is given j and because, by the hypothesis,; the DATA. 41 « * 9. dat. the ratio of the parallelogram AC to I'.G i> given, and that AC is equaP to BL ; therefore the ratio of BL to EG is given - ^^S I and because BL is equiangular to EG, and by the hypothesis, the ratio of BC to FG is given ; thereforg*^ the ratio of K B to « 65. dn. EF is given, and the ratio of KB to BA is given ; the ratio there- fore"* of AB to EF is given. The ratio of AB to £F may be found thus ; Take the straight line MN given in position and magni- tude ; and make the angle MN O equal to the given angle BAK, and the angle MNO equal to the given angle EFG, or AKB : And because the parallelogram BL is equiangular to EG, and has a given ratio to it, and that the ratio of BC to FG is given ; find by the 65th dat. the ratio of KB to KF : and make the ratio of NO to (J? the same with it; Then the ratio of AB to EF is the same with the ratio of MO to OP : For since the tri- an»l« ABK is equiangular to MON, as AB to BKj so is Mb to ON : And as KB to EF, so is NO to OP j therefore exacquali, as AB to EF, so is MO to OP. PROP. LXVIL 70. JLF the sides of two equiangular parallelograms See n. have given ratios to one another; tl\e parallelo- grams shall have a given ratio to one another. h Let ABCD, FFGH be twd equiangular parallelograms, and let the ratio of AB to FF, as also the ratio of BC to FG, be given ; the ratio of the parallelograrn AC to EG is given. Take a straight line K given in magnitude, and because the ratio of AB to EF is given make the ratio of K to L the same with it -, therefore L is given*: And because the ratio of BC to FG is given, make the ratio of L to M the same : Therefore M is given*, and K is given; wherefore** the ratio of K to M is given : But thb parallelogram AC is to the ^ i ^at parallelogram EG, as the straight line K to the straight line M, as • 2. dat. 412 EUCLID'S as is demonstrated in the 23d Prop, of B. 6. Elem. therefore the ratio of AC to EG is given. From this it is plain how the ratio of two equiangular paral- lelograms may be found when the ratios of their sides are given. TO. PROP. LXVIII. SoeN. » 43. dat. «> 9. dat. «67,dat. ^ 35. 1. XF the sides of two parallelograms which have un- equal, but given angles, have given ratios to one another ; the parallelograms shall have a given ratio to one another. Let two parallelograms ABCD, EFGH which have the given unequal angles ABC, EFG have the ratios of their sides, viz. of AB to EF, and of BC to FG, given ; the ratio of the parallelogram AC to EG is given. At the point B of the straight line BC, make the angle CBK equal to the given angle EFG, and complete the parallelo- gram KBCL : And because each of the angles BAK, BKA is given, the triangle ABK is given* inspecics : Therefore the ratio of A B to BK is given ; and the ratio of AB to EF'is given, wherefore^ the ratio of BK to EF is given. And the ratio ofBC to FG is given; r t^ 17 ft and the angle KBC is equal &Jk =Ll_1> KJkl to the angle EFG ; there- fore*^ the ratio of the paral- lelogram KC to EG is given : But KC is equal'^ to AC ; therefore the ratio of AC to EG is given. The ratio of the parallelogram AC to EG may be found thus : Take the straight line IVIN given in positior and mag- nitude, and make the angle MNO equal to the given angle K AB, and the angle NMO equal to the given angle AKB or FEH : And because the ratio of AB to EB" is given, make the ratio of NO toP the same ; also make the ratio of P to Q_the same with the given ratio of BC toFG, the parallelogram AC is to EG, as MO to Q; Because the angle KAB is equal to the angle MNO, and the angle AKB equal to the angle NMO j the triangle AKB is equiangular to NMO : Therefore as KB to B A, so is MC9 to ON i and as BA to EF, so is NO to P j wherefore, ex aequall, as KB to EF, so is MO to P: And BC is to FG, as P 3 to DATA. 413 to CL) and the parallelograms KC, EG are equiangular ; there- fore, as was shewn in Prop. 67, the parallelogram KC, that is, AC is to EG, as MO to Q. Cor. I. If two triangles ABC, DEF have two equal an- "'- gles, or two unequal, but given angles ABC, DEF, and if the ratios of the sides about these angles, . _, viz. the ratios of A B to DE, and of ^ Y D H BC to EF be given ; the triangles shall have a given ratio to one ano- ther. 3 c g f Complete the parallelograms, BG, EH ; the ratioof BGto EH is given^ ; and therefore the tri- .ejorf-i. angles which are the halves'^ of them have a given' ratio to h 34''", one another. " '^- ^• Cor. 2. If the bases BC, EF of two triangles ABC, E)EF '^• have a given ratio to one another, and if also the straight lines AG, DH which are drawn to the bases from the opposite an- gles, either in equal angles, or unequal, but given angles AGO, DHF have a given ratio to one j^ j^ L D another; thetrianglesshall have a given ratio to one another. Draw BKjELparallel to AG, / ti DH, and complete the paral-'-> Vx C jlL. Jfi F lelograms KC, LF. and because the angles AGC, DHF, or their equals, the angles KBC, LEF are either equal, or un- equal, but given ; and that the ratio of AG to DH, that is, of KB to LE, is given, as also the ratio of BCto EF ; therefore* ,67 or 68. the ratio of the parallelogram KC to LF is given ; wherefore '* also the ratio of the triangle ABC to DEF is giveIl^ "^tl 3" 61. PROP. LXIX. i F a parallelogram which has a given angle he ap- plied to one side of a rectilineal figure given in spe- cies ; if the figure have a given ratio to the parallelo- gram, the parallelogram is given in species. Let ABCD be a rectilineal figure given in species, and to one side of it AB, let the parallelogram ABEF having the given angle ABE be applied ; if the figure ABCD hasagiven ratio to the parallelogram BF, the parallelogram BF is given in species. Through the point A draw AG parallel to BC, and through the point C draw CG parallel to AB, and produce GA, CB to the 4H »i3. cief. » 53. ciat. « 9. Haf. ".3,5. 1. « 1. 6, E U C L I D'S the points H, K ; because the angle ABC Is given*, and the ratioof ABto BC is given, the figure ABCD being given in species ; therefore, the parallelogram BG is given* in species. And b..'cause upon the same straight line AB the two rectilineal Htiures BD, BG given in species are described, the ratio of BD to BG is given'' ; and, by hypothesis, the ratio of BD to the parailelogram BF is given ; wherefore*' the ratio of BF, that is'', of ths'paralleloi;ram BH, to BG is given, and therefore^ the ratio of the straight line KB to BC is given ; and the ratio of BC to BA is given, wherefore the ratio of KB to BA is given'' : And because the angle ABC is given, theadJ4centangle ABKisgiven ; and the angle ABE is jriven, there to|e the remaining angle KBEis given. The angle EKB is also given, bece ise it is equal to the angle ABK ; theretore the triafigle. BKE is given in species, and consequently the ratio of EB to EK is given j and the ratio of KB to BA is giv^n wherefore'^ the latio or j^f. EB to BA is given ; and the angle ABE is given, therefore the parallelo- gram BF is given* in species. A parallelogram simi- lar to Bt may be found thus : Take a straight line LM given in position and magnitude^ and because the angles ABK, ABE are given, mj^ke the angle NLM equal to ABK, and the angle NI.O equal to ABE. And because the ratio of BF to BD is given, ma.ke tne ratio of LM to P the same with it ; and because the ratio of the figure BD to BG is given, find this ratio i)y the 53d dat. and make' the ratio of P to Q the same. Also, because the ratio of CB to BA is given, make the ratio of Qjo R the same ; and take LN equal to R ; through the point ivJ draw OM parallel to LN, and complete the parallelogram N LOS j then this is similar tpthe parallelogram BY. . ' Becausethe angle ABK is equal to NT>M,ah(rthe angle ABE to NLO, file angle KHE is equal to MLOj and the angles BKE, LMO are equal, because the angle ABK is equal to NLM; therefore the triangles l^KE, LMO are equiangu- lar to one another ; wherefore, as BE to BK, so is LO to LM ; and because as the figure BF to BD, so is the straight line LM to P; and as BD to BG, so is P to Qj ex aequali, as BF, that is**, BPI to BG, so is LM to Q ; brut EH is to^ BG, DATA. 415 BG, as KB to BC ; as therefore KB to BC, so is LM to Qj and because BE is toBK as LO to LM ; and as BK to BC, so is LM to Q^: And as BC to BA, so Q^was made to R ; therefore, ex sequali, as BE to BA, so is LO to R, that is, to LN ; and the angles ABE, NLO are ecjual ; therefore the parallelogram BF is similar to LS. PROP. LXX. 62. 7». JLF two straight lines have a given ratio to one ano- See n. ther, and upon one of them be described a rectili- neal figure given in species, and upon the other a parallelogram having a given angle; if the figure have a given ratio to the parallelogram, the paral- lelogram is given in species. Let the two straight lines AB, CD have a given ratio to one another, and upon AB let the figure AEB given in species be described, and upon CD the parallelogram DF having th« given angle FCD ; if the ratio of AEB to DF be given, the parallelogram DF is given in species. ' Upon the straight line AB, conceive the parallelogram AG to be described similar, and similarly placed to FD ; and because the ratio of AB to CD is given, and upon them are described the similar rectilineal figures AG, FD i the ratio of AG to FD is given*) and the ratio of FD to AEB is given ; therefore '' the ratio of AEB to AG is given ; and the angle ABG is given, because it isequal to the angle FCD j because therefore the parallelogram AG which has a given angle ABG is applied to a side AB of the figure AEB given in species, and the ratio of AEB to AG is given, the parallelo- gram AG is given'' in species ; but FD is similar to AG j «: 69. dat. therefore FD is given in species. A parallelogram similar to FD may be found thus : Take a straight line H given in magnitude ; and because the ratio of the figure AEB to FD is given, make the ratio of H to K the same with it : Also, because the ratio of the straight line CD to AB is given, find by the 54th dat. the ratio which the figure FD described upon CD has to the figure AG described upon AB similar to FD j and make the ratio of K to L the same with this ratio : And because the ratios of H to K, and of K to 4i6 EUCLID'S i> p. dat. to L arc given, the ratio of H to L is given^ j because, the re- fore, ?s AEB to FD, so is H to K : and as FD to AG, so is K to L i ex lequali, as AEB to AG so is H to L ; therefore the ratio of AEB to AG is given ; and the figure AEB is given in species, and to its side AB the parallelogram AG is applied in the given artgle ABG ; therefore by tne 69th dat. a parallelo- gram may be found similar to AG : Let this be the parallelo- gram MN ; MN also is similar to FD ; for, by the construc- tion, MN is similar to AG, and AG is similar to FD; there- fore the parallelogram FD is similar to MN. PROP. LXXI. ^*- 1_F the extremes of three proportional straight lines have given ratios to the extremes of other three proportional straight hnes ; the means shall also have a given ratio to one another: And if one extreme has a given ratio to one extreme, and the liiean to the mean ; likewise the other extreme shall have to the other a given ratio. Let A, B, C be three proportional straight lines, and D, E, F, three other ; and let the ratios of A to D, and of C to F, be given J then the ratio of B to E is also given. Because the ratio of A to D, as also of C to F, is given, the » 67. dat. ratio of the rectangle A, C to the rectangle D, F' is given*; b n. 6. but the square of B is equal"" to the rectangle A, C ; and the square of £ to the retflangle'' D, F ; therefore the ratio of the «58. daf. square of B to the square of E is given ; wherefore"^ also the ratio of the straight line B to E is given. Next, let the ratio of A to D, and of B to E, be given ; then the ratio of C to F is also given. -' *■'*'' Because the ratio of B to E is given, the ratio of l-k X * 54. Jat. -^^g square of B to the square of E is given'' ; there- 1~~* fore'' the ratio of the redlangle A, C to the redangle Y 1 D, F is given; and the ratio of the side A to the | side D is given ; therefore the ratio of the other ( • e ■. flat, sjjjg Q jQ t)^e other F is given''. Cor. And if the extremes of four proportionals have to the extremes of foijr other proportionals given ratios, and one of the means a given ratio to one of the means ; the other mean shall have a given ratio to the other mean, as may be shewn in the same mamier as in the foregoing proposition. DATA. 417 PROP. LXXII. 1 F four straight lines be proportionals; as the first is to the straight line to which the second has a given ratio, so is the third to a straight line to which the fourth has a given ratio. Let A, B, C, D be four proportional straight lines, viz. as A to B, so C to D ; as A is to the straight line to which B has a given ratio, so is C to a straight line to which D has a given ratio. • Let E be the straight line to which B has a given ratio, and as B to E, so make D to F : The ratio of B to E is given*, and therefore the ratio of D to F ; and because as A to B, so is C to D ; and as B to E, . $0 D to F ; therefore, ex aequali, as A to E, so is A. x> xl C to F ; and E is the straight line to which B has a C D F given ratio, and F that to which D has a given ratio ; therefore as A is to the straight line to which B has a given ratio, so is C to a line to which D has a given ratio. S2. •H> PKOP. lxxiil 83. J.F four straight lines be proportionals ; as the first seeN. is to the straight line to which the second has a given ratio, so is a straight line to which the third has a given ratio to the fourth. Let the straight line A be to B, as C to D ; as A to the straight line to which B has a given ratio, so is a straight line to which C has a given ratio to D. Let E be the straight line to which B has a given ratio, and as B to E, so make F to C ; because the ratio of B to E is given, the ratio of C to F is given : And because A is to B, as C to D ; and as B to E, so F to C : therefore, ex aequali, in pro- portLone perturbato", A is to E, as F to D ; that is, | • J3. 5. A is to E to which B has a given ratio, as F, to which C has a given ratio, is to D. Ee A BE FCD 4i8 E U C L I D'S Si. PROP. LXXIV. J F a triangle has a given obtuse angle ; the excess ot" the square ot the side which subtends the obtuse angle, above the squares of the sides which contain it, shall have a given ratio to the triangle. Let the triangle ABC have a given obtuse angle ABC j and produce the siraight line CB, and £rom the point A draw AD perpendicular to BC : The excess of the square of AC above ».12 2. the squares of AB, BC, that is% the double of the rectangle contained by DB^ BC, has a given ratio to the triangle ABC. Because theangle ABCis given, theangle A BD is also given ; and the angle ADB is given ; wherefore the triangle ABD "43. «iat. is given'^ in species ; and therefore the ratio of AD to DB M. 6. is given : And as AD to DB, so is<^ the re£langle AD, BC to the rectangle DB, BC ; whercFore th» ratio of the re6^- angle AD, BC to tKe reiiangle DB, BC is given, as also the ratio of twice the rectan^j^le DB, BC to the re6langle AD, BC : But the ratio of the i entangle A D, BC to the triangle ABC "41.1. is given, because it is double** of the tri- angle ; therefore the ratio of twice the redtangle DB, BC to the triangle ARC is "^^.dat. given"^; and twice the redlangle-DB, BC IS the excess^ of the square of AC above the squares of AB, BC J therefore this excess has a given ratio to the triangle ABC. And the ratio of this excess to the triangle ABC naay be found thus: Take a straight line EF given in position and mag- nitude i and because the angle ABC is given, at the point F of the strai:;ht line F,F, make the angle EFG equal to the ang|e ABCj produce GF, and draw EH perpendicular to FG; then the ratio of the excess of the square of AC above the squares of AB, BC to the triangle ABC, is the same with the ratio of quadruple the straight line HF to HE. Because the angle ABD is equal to the angle EFH, and the angle ADB to EHF, each being a right angle : tjje tri- '4.6. angle ADB is equiangular to EHF; therefore*" as BD to DA, «Cor. 4.5. so FH to HE; and as quadruple of BD to DA, so is 8 qua- druple of B'H to HE : Biit as twice BD is to DA, so is*^ twice the redangle DB, BC o the redangle AD, BC ; and as DA T. 5. to the half of it, so i.s'' the rectangle AD, BC to its half the triangle C5. DATA. triangle ABC ; therefore, ex aequali, as twice BD is to the half of DA, that is, as quadruple of BD is to D.A, that is, as qua- druple of FH to HE, so is twice the reJlaiigle DB, BC to the triangle ABC. PROP. LXXV. 1 F a triangle has a given acute angle, the space by which the square of the side subtending the acute angle is less than the squares of the sides which con- tain it, shall have a given ratio to the triangle. Let the triangle ABC have a given acute angle ABC, and draw AD perpendicular to BC, the space by which the square of AC is less than the squares of AB, BC, that is% the double » 13. 2. of the rectangle contained by CB, BD, has a given ratio to the triangle ABC. Because the angles ABD, ADB are each of them given, the triaagle ABD is given in species ; and therefore the ratio of BD to DA is given : And as BD to DA, so is the re<5tangle CB, BD to the re<Ebng!e CB, AD : Therefore the ratio of these re<£l- ang!cs is given, as also the ratio of twice the reaangle CB, BD, to the reclangle CB, AD, but the rectangle CB, AD has a given ratio to its half the triangle ABC : Therefore'' the ratio of twice the redanglc CB, BD to the triangle ABC is given : and twice the rectangle CB, BD is* the space by which the square of AC is less than the squares of AB, BC ; there- fore the ratio of this space to the triangle ABC is given : And the ratio may be found as in the preceding proposition. LEMMA. I F from the vertex A of an isosceles triangle A BC, any straight line AD be drawn to the base BC, the square of the side AB is equal to the rectangle BD, DC of the segments of the base together with the square of AD ; but if AD be drawn to the base produced, the square of AD is equal to the rectangle BD, DC, together with the square of AB. Case i. Bisect the base BC in E, and join AE, which will be perpendicular* to BC J wherefore the square of AB is equal *• to the squares of AE, EB; but the square of EB is equal* to the re«angle BD, DC together with the square of DE ; there- fore the square of AB is equal to the E e 2 squares 419 DC 9. dat »8. 1. ^47. 1. \2<5 E U C t I D ' S MT. ;. squares of AE, fcD, that is, to^ the square of AD, together with the rectangle BD, DCj the other case is shewn in the same way by 6. 2. Elem. ej. PROP. LXXVl. xF a triangle have a given angle, the excess of the square of the straight line which is equal to the two sides that contain the given angle, above the square of the third side, shall have a given ratio to the triangle. / Let the triangle ABC have the given angle BAC, the excess of the square oT the straight line which is equal to BA, AC together above the square of BC>.-shalI have a given ratio to the triangle ABC. Produce BA, and take AD equal to AC, ioin DC, and produce it to E, and through the point B draw BE parallel to AC; join AE, and draw AF perpendicular to DC ; and be- cause AD is equal to AC, BD is equal to BE ; and BC is drawn from the vertex B of the isosceles triangle DBE ; there- fore, by the Lemma, the square of BD, that is, of BA and AC together, is equal to the rectangle DC, CE together with the square of BC ; and therefore the square of BA, AC to- gether, that is, of BD, is greater than the square of BC by the rectangle DC, CE ; and this rectangle has a given ratio to the triangle ABC, because the angle BAC is given, the adjacent angle CAD is given ; and each of the , angles ADC, DCA is given, for •5. &: 32. each of them is the half^ of the given ^^^^^^_\ '• angle BAC; therefore the triangle ^ \~!L *43. dat.' ADC is givent> in species ; and AF is EL drawn from its vertex to the base in ■^ a given angle ; wherefoie the ratio of AF to the base CD is * 50. dat. given" and as CD to AF, so is"^ the rectangle DC, CE to " 1- <3- the rectangle AF, CE ; and the ratio of the rectangle AF, '41.1. CE to its hall"=, the triangle ACK, is given; therefore the ' '". I. ratio of the rectangle DC, CE to the triangle ACE, that is', « 9, dat, to the triangle ABC, is given? ; and the rectangle DC, CE is the excess of the square of BA, AC, together above the square of BC : Therefore the ratio of this excess to the tri- angle ABC is given. The ratio which the rectangle DC, CE, has to the triangle ABC is found thus : Take the straight line HG given in posi- tion 6. no 5. DATA. 4H tion and magnitude, and at the point G in GH make the angle HGK equal to the given angle CAX>, and take GK equal to GH, join KH, and draw GL perpendicular to it : Then the ratio of HK to the half of GL is the same with the ratio of the rectangle DC,CE to the triangle ABC : Beouse the angles HGK, DAC, at the vertices of the isosceles triangles GHK, ADC, are equal to one another, these triangles are similar i and because GL, AF, are perpendicular to the bases HK,DC, as HK to GL, so is ^ (DC to AF, and so is) the rectangle "> i^ DC, C£ to the rectangle AF, CEj but as GL to its half, so is the rectangle AF, CE to its half, which is the triangle ACE, or the triangle ABC;- therefore, ex acquaii, HK is to the half of the straight Line GL, as the rectangle DC, C£, ir to the triangle ABC. - •- Cor. And if a triangle have a given angle, the spacE by which the square of the straight line, which is the difference of the sides which contain the given angle, is less than the square of the third side, shall have a given ratio to the trian- gle. This is demonstrated the same way as in the preceding proposition, by help of the second case of the Lemma. PROP. LXXVIL I. If the perpendicular drawn from a given angle of as*eN. triangle to the opposite side, or base, has a given ratio to the base, the triangle is given in species. Let the triangle ABC have the given angle BAC, and let the perpendicular AD drawn to the base BC, have a given ratio to it, the triangle ABC is given in species. If ABC be an isosceles triangle, it is evident*, that if any 5.1c 22. 1. A B Pv D C one of its angles be given, the rest are also given ; and there- fore the triangle is given in species, without the consideration ©f the ratio of the perpendicular to the base, which in this case is given by Prop. 50. But when .-VBC is not an isosceles triangle, take any straight line EF given in position and magnitude, and upon it descr-.be E e 3 the 422 E U C LI D'S the segment of a circle EGF, containing an angle equal to thf given angle BAC, draw GH bisecting EF at right angles, and join EG, OF : Then, since the angle EGF is equal to the angle BAC, and that EGF is an isosceles triangle, and ABC is not the angle FEG is not ifequal to the angle CBA : Draw EL making the angle FEL equal to the angle CBA ; join FL, and drawLM perpendicular to EF; then because the triangles ELF BAC are equiangular, as also are the triangles MLE, DAB, as ML to LE, so is DA to AB ; and as LE to EF, so is A B to BC; wherefore, ex aequali, as LM to EF, so is AD to BC ; and because the ratio of AD to BC is given, therefore the ratio of LM to EF is given ; and EF is given, wherefore^ LM also is given. Complete the parallelogram LMFK ; and because LM is given, FK is given in magnitude; it is also given in position; and the point F is given, and consequently*^ the point K ; and because through K the straight line KL is drawn parallel to EF, " 3J. da', which is given in position, therefore** KL is given in position : 2. dat. « 30. da' <= 28. dat. ♦ -19. dat. « 42. dat. K D C and the circumference ELF is given in position ; therefore the point L is given^. And because the points, L, E, F, are given, the straight lines LE, EF\ FL,are given*^in magnitude ; there- fore the triangle LEF is given in species 8 ; and the triangle ABC is similar to LEF, whereforealso ABC isgiven inspecies. Because LM is less than GH, the ratio of LM to EF, that is, the given ratio of AD to BC, must be less than the ratio of GH to EF, which the straight line, in a segment of a circle containing an angle equal to the given angle, that bisects the base of the segment at right angles, has unto the base. Cor. I. If two triangks, ABC, LEF have one angle BAC equal to one angle ELF, and if the perpendicular AD be to the base BC, as ihe perpendicular LM to the base EF, the trian- gles ABC, LEF are similar. Describe the circle EGF about the triangle ELF, and draw LN parallel to EF, join EN, NF, and draw NO perpendicu- lar to EF i because the angles ENF, ELF are equal, and that the DATA. 423 the> angle EFN is equal to the alternate angle FNL, that is, to the angle FEL in the same segment ; therefore the triangle N£F is similar to LEF; and in the segment EGF there can be no other triangle upon the base KF, which has the ratio of its perpendicular to that base the same w th the ratio of LM or NO to EF, because the perpendicular must be greater or less than LM or NO ; but, as has been shewn in the preceding demonstration, a triangle, similar to ABG, can be described in the segment EGF upon the base EF, and the ratio of its per- pendicular to the base is the same, as was there shewn, with the ratio of AD to BC, that is, of LM to EF ; therefore that triangle must be either LEF, or NEF, which therefore are similar to ihe triangle ABC. Cor. 2. If a triangle ABC has a given angle BAC, and if the straight line AR drawn from the given angle to the op- posite side BC, in a given angle ARC, has a given ratio to BC, the triangle ABC is given in species. » Draw AD perpendicular to BC ; therefore the trianj^lc ARD is given in species j wherefore the ratio of AD to AS. is given : and the ratio of AR to BC is given, and consequently'' " 9. dat. the ratio of AD to BC is given; and the triangle ABC is therefore given in species'. '77.dat. Cor. 3. If two triangles ABC, LEF have one angle BAC equal to one angle ELF, an ! if straight lines drawn from these angles to the bases, making with chem given and equal angles, have the same ratio to the bases, each to each ; then the triangles are similar; for having drawn perpendiculars t« the bases from the equal angles, as one perpendicular is to its base, so is the other to its base^j wherefore, by Cor. i. the "5 22. 5. triangles are similar, A triangle similar to ABC may be found thus : Having de- scribed the segment EGF, and drawn the straight line GH as was dlre£led m the j^oposition, find FK, which has to EF the given ratio of AD to BC ; and place FK at right angles to EF from the point F ; then because, as has been >hewn, the ratio of AD to BC, that is, of FK to EF, must be less than the ratio of GB to EF ; therefore FK is less than G H ; and consequently the parallel to EF drawn through the point K, must meet the circumference or the segment in two points: Let L be either of them, and join EL, LF, and draw LM perpendicular to EF : then, because the angle BAC is equal to the anijle ELF, and that AD ts^to BC, as KF ; that is, LM to EF, The triangle ABC is similar to the triangle LEF, by Cor. i. E 64 424 fe U C L I D'S 80. '41. 1. ' Cor. 62. dat. 9. dat. '1.6. 77. dat. PROP. LXXVIII. IF a triangle have one angle given, and if the ra- tio of the rectangle of the sides which contain the given angle to the square of the third side be given, the triangle is given in species. Let the triangle ABC have the given angle BAC, and let the ratio of the reftangle BA, AC to the square of BC be given J the triangle ABC is given in species. From the point A, draw AD perpendicular to BC, the re6^- angle AD, BC has a given ratio to its half* the triangle ABC j and becausfe the angle BAC is given, the ratio of the triangle BAC to the rectangle B A, AC is given^ j and by the hypo- thesis, the ratio of the redlangle BA, AC to the square of BC is given j therefore' the ratio of the redangle AD, BC to the square of BC, that is*^, the ratio of the straight line AD to BC is given ; wherefore the triangle ABC is given in species'^. A triangle similar to ABC may be found thus : Take a straight line EF given in position and magnitude, and make the angle FEG equal to the given angle BAC, and driaw FH perpendicular to EG, and BK perpendicular to AC : therefore thetrianglesABKjEFH are similar, and the reft- a an2;le AD, BC or the reaangle BK,AC which is equal to it, is to the rectangle B A, AC as the straight line BK to BA, ii K thatis,asFHtoFE. Let the given ratio of the rectangle B A, AC to the square of BC be the same with the ratio of the straight line EF to FL j there- fore, ex aequali, the rjrtio of the reftangle AD, BC to the square of BC, that is, the ratio of the straight line AD to BC, is the same with the ratio of HF to FL ; and because AD is not greater than the straight line MN in the segment of thfe circle described about the triangle ABC, whicli bisects BC at right angles ; the ratio of AD to BC, that is, of HF to FL, must not be greater than the ratio of MN to BC : Let ic be so; and, by the 77th dat. find a triangle OPQ which has one of its angles POQ^ equal to the given angle BAC, and the ratio of the perpendicular OR, drawn from that angle to the base PQ, the same with the ratio of HF to FL ; then the triangle ABC is similar to OPQ 10. dat DATA. 4^5 OPO : Because, as has been shewn, the ratio of AD to BC is the "same with the ratio of (HF to FL, that is, by the con- struction, with the ratio of) OR to PQ; ajid the angle BAG is equal to the angle POQ. Therefore the triangle ABC is similar^ to the trianele POQ. *^i. Cot. OthtrxeisCy Let tht triangle ABC have the given angle BAC, and let the ratio of the rectangle BA, AC tto the square of BC be giv^en i the triangle ABC is given in species. Because the angle BAC is given, the excess of the square of both the sides BA, AC together above the square of the third side BC has a given* ratio to the triangle ABC. Let the »76. dat. figure D be equal to this excess ; therefore the ratio of D to the triangle ABC is given ; and the ratio of the triangle ABC to the rectangle BA, AC is sriven\ because BAC is a siven „ angle ; and the redangle BA, AC has ^ ] ] ^^^^' a given ratio to the square of BC; wherefore*^ the ratio of D to the square of BC is given ; and, by cotn- position<*, the ratio of the space D, i^ ^ «7 dat. together with the square of BC to the square of BC is given; but D together with the square of BC is equal to the square of both BA and AC toother ; therefore the ratio ofthe square of BA, AC together to the square of BC is given ; and the ratio of B A, AC together to BC is therefore given'' ; and the angle BAC is given, wherefore^ the triangle ABC is given in species. ' ^^' ^*- The composition of this, which depends upon those of the ^^' ^^^ 76th and 48th propositions, is more complex than the preced- ing composition, which depends upon that of Prop. 77. which is easy. PROP. LXXIX. K. IF a triangle ])ave a given angle, and if the straight see x. hne drawn from that angle to the base, making a given angle with it, divides the base into segments which have a given ratio to one another ; the trian- gle is given in species. . • Let the triai>gle ABC have the given angle BAC, and let the straight line AD drawn to the base BC, making the given angle ADB, divide CB in;? the segments BD, DC which hkve '47.dat, 4i6 E U C L 1 D'S a given ratio to one another ; the triangle ABC is given in species » 5. 4. Describe* the circle BAG about the triangle, and from its centre E, draw EA, EB, EC, ED ; because the angle BAG is «» 20^3. given, the angle BEC at the centre, which is the double'' of it, is given. And the ratio of BE to EC is given, because they *4i. dat. are equal to one another; therefore the<= triangle BEC is given in species, and the ratio of EB to BC is given ; also the « 7.dat. ratio of CB to BD is given"*, because the ratio of BD to DC «9, dat. is given ; therefore the ratio of EB to BD is givcn%*and the angle EBC is given, wherefore the triangle EBD is given* in species, and the ratio of EB, that is, of EA, to ED, is there- fore cfiven ; and the angle EDA is given, because each of the angles BDE, BDA is given ; therefore the triangle AKD is given'' in species, and the angle AED given ; also the angle DEC is given, be- cause each of the angles BED, BEC is given ; therefore the angle AEC is given, and the ratio of EA to EC, which are equal, is given ; and the triangle AKC is 1^\ therefore given^ in species, and the angle ECA is given ; and the angle ECB is given, wherefore the angle ACB is given, and the angle BAC is also given i therefore? the triangle ABC is given in species. A triangle similar to ABC may be found, by taking a straight line given in position and magnitude, and dividing it in the given ratio which the segments BD, DC are required to have to one another ; then, if upon that straight line a segment of a circle be described containing an angle equal to the given angle BAC, and a straight line be drawn from the point of division in an angle equal to the given angle ADB, and from the point where it meets the circumference, straight lines be drawn to the ■extremity' of the first line, these, together with the first line, shall contain a triangle similar to ABC, as may easily be shewn. The demonstration may be also made in the manner of that of the 77th Prop, and that of the 77th may be made in the manner of this. PROP. LXXX. XF the sides about an angle of a triangle have a gi- ven ratio to one auotlier, and if the perpendicular drawn from tliat angle to the base has a given ra- tio to the base ; the triangle i; given in species. Lst « 43.*dat. DATA. 427 Let the sides BA, AC, aboQt the angle BAG of the trian- gle ABC/ have a given ratio to one another, and let the per- pendicular AD have a given ratio to the base BC, the triangle ABC is given in species. First, let the sides AB, AC be equal to one another, there- fore the perpendicular AD bisects- the base *26. 1, BC; and the ratio of AD to BC, and there- ^ fore to its halt DBy is given ; and trie angle < / \ ..^ ADB is given; wherefore the triangle* ABD y. \ •'^*'- and consequently the triangle ABC is given'' p rv ~p '144.dat, in species. But let the sides be unequal, and BA be greater than AC ; and malce the angle CAE equal to the angle ABC ; because the angle AEB is common to the triangleb AEB, CEA, they are similar ; therefore as AB to BE, so is CA to AE, and, by permutation, as BA to AC, so is BE to EA, and so is EA to EC ; and the ratio of BA to AC is given, therefore the ratio of BE to E A, and the ratio of EA to EC, as also the ratio of BE to EC is given*"; wherefore the ratio of EB tOcp ,jat- BCis given"*; and the ratio of AD to BC "6 dzt, is given by the hypothesis, therefore'^ the -^ ratio of AD to BE is given ; and the ratio -^ • of BE to EA was shewn to be given ; whereforetheratioof AD toEA is given; ^ [," ^ Y, E' and ADE is a right angle, therefore the triangle ADE is given' in species, and the angle AEB given; the ratio of BE to EA is likewise given, therefore*" the trian- *'^- *■'**• gle ABE is given in species, and consequently the angle E AB, as also the angle ABE, that is, the angle CAE is given; therefore the angle BAG is given, and the angle ABC being also given, the triangle ABC is given* in species. '^43. dat. How to find a triangle which shall have the things which are mentioned to be given in the proposition, is evident in the first case ; and to find it the more easily in the other case, it is to be observ ed that, if the straight line EF equal to EA be placed in EB towards B, the point F divides the base BC into the segments BF, FC, which have to one another the ratio of the sides BA, CA, because BE, EA, or EF, and •19. 5. EC, were shown to be proportionals, therefore* BF is to FC, as BE to EF, or EA, that is, as BA to AC ; and AE cannot be less than the altitude of the triangle ABC, but it may be equal 4«8 E U C L I D'S equal to it, which, if it be, the triangle, in this case, as also the ratio of the sides, may be thus found : Having given the ratio of the perpendicular to the base, take the straight line • GH, given in position and magnitude, for the base of the tri- angle to be found; and let the given ratio of the perpendicular to the base be that of the straight line K to GH, that is, let K be equal to the perpendicular ; and suppose GLH to be the triangle which is to be found, therefore having made the angle HLM equal to LGH, it is required that LM be perpen- dicular to GM, and equal to K; and because GM, ML, I MH are proportionals, as was shewn of BE, EA, EC, the re<9angle GMH is equal to the square of ML. Add the com- mon square of NH (having bisected GH in N) and the square «6.2. of NM is equal* to the squares of the given straight lines NH and ML, or K i therefore the square of NM, and its side NM is given, as also the point M, viz. by taking the straight line NM, the square of which is equal to the squares of NH, ML. Draw ML equal to K, at right angles to GM ; and be- cause ML is given in position and magnitude, therefore the point L is given, join LG, LH ; then the triangle LGH is that which was to be found ; for the square of NM is equal to the squares of NH and ML, and takmg away the common square of NH, the re6l- angle GMH is equale to the square of ML ; there- fore as GxM to ML, so is ML to MH, and the tri- *€.6. angle LGM is'' therefore equiangular to HLM, and the angle HLM equal to Q the angle LGM, and the straight line LM, drawn from the vertex of the triangle mak- ing the angle HLM equal to LGH, is perpendicular to the base and equal to the given straight line K, as was required ; and the ratio of the sides GL, LH, is the same with the ratio of GM to ML, tKat is, with the ratio of the straight line which is made lip of GN, the half of the given base and of NM, the sqiiare of which is equal to the squares of GN and K, to the straight ^lin€ JC. And whether this ratio of GM to ML is greater or )ess than the ratio of the sides of any other triangle upon the basft GH, and of y^hich the altitude is equal to the straight line K, 3 that NilR KF DATA. ._ 429 that is, the vertex of which is in the parallel to GH drawn through the point L, may be thus found. Let OGH be any such triangle, and draw OP, making the angle HOP equal to the angle OGH ; therefore, as before, GP, PO, PH are pro- portiona,ls, and PO cannot be equal to LM, because the rect- angle GPH would be equal to the rectangle GMH, which is impossible; for the point P cannot fall upon M, because O would then fall on L ; nor can PO be less than LM, therefore it is greater ; and consequently the rectangle GPH is greater tlan the rectangle GMH, and the straight line GP greater than GM : Therefore the ratio of GM to MH is greater than the ratio of GP to PH, and the ratio of the square of GM to the square of ML is therefore' greater than the ratio of' 2. Cct. the square of GP to the square of PO, and the ratio of the "^ ^' straight line GM to ML greater than the ratio of GP to PO. But as GM to ML, so is GL to LH ; and as GP to PO, so is GO to OH ; therefore the ratio of GL to LH is greater than the ratio of GO to OH ; wherefore the ratio of GL to LH is the greatest of all others; and consequently the given ratio of the greater side to the less must not be greater than this ratio. But if the ratio of the sides be not the same with this greatest ratio of GM to ML, it must necessarily be less than it : Let any less ratio be given, and the same thmgs being supposed, viz. that GH is the base, and K equal to tne altitude of the triangle, it may be found as follows: Divide GH in the point Q, so that the ratio of GQ to QH may be the same with the ^iven ratio of the sides ; and as GQ to QH, so make GP to f Q and so wilK PQ b?" to PH ; wherefore the square of GP is to the square of FQ, as« the straight line GP to PH : ♦ 19. 5. And because GM, ML, MH are proportionals, the square of GM is to the square of ML, as' the straight line GM to MH : But the ratio of GQ_to Q.H, that is, the ratio of GP to PQ, is Ie?s than the ratio of GM to ML ; and therefore the ratio of the square of GP to the square of PQ^is less than the ratio of the square of GM to that of ML; and consequently the ratio of the straight line GP to PH is less than the ratio of GM to MH ; and, by division, the ratio of GH to HP is less than that of GH to HM ; wherefore'' the straight line HP is k 10. 5, greater than HM, and the reftangie GPH, that is, the square of PQ, greater than the reiftangle GMH, that is, than the square of ML, and the straight line PQ^is therefore greater than 43° E U C L I D'S thaa M!^. Draw LR parallel to GP, and from P draw PR zt right angles to GP. Because PQ_is greater than ML, or PR, the circle described from the centre P, at the distance PQ, must necessarily cut LR in two points ; let these be O 8, and join OG, OH ; SG, SH : each of the triangles OGH, ^GH have the things mentioned to be given in the proposition : Join OP, SP J and because as GP to PQ, or PO, so is PO to PH, the triangle OGP is equiangular to HOP ; as, there- fore, OG to GP, so is HO to OP ; and, by permutation, as GO to OH, so is GP to PO, or PQ_; and so is GQ.to QH ; Therefore the triangle OGH has the ratio of its sides GO, OH, the same, with the given ratio of GCi to QH : and the perpen- dicular has to the base the given ratio of K to GH, because the perpendicular is equal to LM, or K : The like may be shewn m the same way ot the triangle SGH. This construction by which the triangle OGH is found, is shorter than that which would be deduced from the demon- stration of the datum, by reason that the base GH is given in position and magnitude, which was not supposed in the de- monstration : The same thing is to be observed in the next proposition. M. PROP. LXXXI. XF the sides about an angle of a triangle be une- qual, and have a given ratio to one another, and if the perpendicular from timt angle to the base di- vides it into segments that have a given ratio' to one another, the triangle is given in species. Let ABC be a triangle, the sides of which about the angle BAC are unequal, and have a given ratio to one another, and let the perpendicular AD to the base BC divide it into the segments BD, DC, which have a given ratio to one another, the triangle ABC is given in species. .'" Jjct, AB be greater than AC, and make the angle CAE cqtial to the angle ABC ; and because the angle AEB is com- »4. 6. mon to the triangles ABE, CAE, they are* equiangular to ojie another ; Therefore as AB to BE, so is CA to AE, and, by XLH ^' DATA. by permutation, as AB to AC, so BE to E A, and so is EA to EC : But the ratio of BA to AC is given, therefore the ratio of BE to EA, as also the ratio of EA to EC is given ; wherefore'' the ratio of BE to EC, as also"^ the ratio of EC to CB is given : And the ratio of BC to CD is given% because the ratio of BD to DC is given ^ therefore* the ratio of EC to CD is given, and consequently* the {> ratio of DE to EC : And the ratio of EC to EA was shewn to be given, therefore'' the ratio of DE to EA is given : And ADE is a right angle, wherefore^ the tri- angle A.DE is given in species, and the angle A ED given ; And the ratio of CE to EA is given, therefore^ the triangle AEC is given in species, and consequently the angle ACE is given, as also the adjacentangle ACB. In the same manner, because the ratio of BE to EA is given, the triangle BE A is given in species, a.d the angle ABE is theretore given : And the angle ACB is given ; wherefore the triangle ABC is givens in species. But the ratio of the greater side BA to the other AC must be less than the ratio of the greater segment BD to DC : Be- cause the square of BA is to the square of AC, as the squares of BD, DA to the squares of DC, DA ; and the squares of BD, DA have to the squares of DC, DA, a less ratio than the square of BD has to the square of DCf, because the square of BU is greater than the square of DC j therefore the square of jliA l^s to the square of AC a less ratio than the square of BD has to that of DC : And consequently thi ratio of BA to AC is less than the ratio of BD to DC. This being premised, a triangle w^hich shall have the things mentioned to be given in the projX)sition, and to which the triangle ABC iy similar, may be found thus : Talce a straight line GH given in position and magnitude, and divide it in K, so that the ratio ot GK to KH may be the same with the given ratio of BA to AC : Divide also GH in L, so that the ratio of ♦31 * 46. dat. 'U. dat. « 43. dat. ■}■ K 'i. ';? g-^ilf- (hau B, ir.l O zny :.^.-d magnitude; then A and C together hare to B and C tfjg- ther a l?-! ralio than A has to B. L-: A be to B as C to D, aud because A i* greater than B, C is greitT than I>: But as A i? !o P, so A and C to B and D ; and A and C have to B ani C a less ra- tio than A a-'d C Hnve to B and P, becau* C is greater than D, thereiore A and C h&«e lb 2 *pdC iea ;.itic '.hen A to B. y 432 EUCLID'S of GL to LH may be the same with the given ratio of BD t9 DC, and draw LM at right angles to GH : And because the ratio of the sides of a triangle is less than the ratio of the seg- ments of the base, as has been shewn, the ratio of GK tq KH is less than the ratio of GL to LH j wherefore the point L must fall betwixt K and H : Also make as GK to KH, so GN fciP. 5. toNK, and so shall'^NK be toNH. And from the centre N, at the distance NK, describe a circle, and let its circumference meet LM in O, and join OG, OH j then OGH is the triangle which was to be described; Because GN is to NK, or NO, as NO to N H, the triangle OGN is equiangular to HON j there- fore as OG to GN, so is HO to ON, and, by permutation, as GO to OH, so is GN to NO, or NK, that is, as GK to KH, that is, in the given ratio of the sides, and by the con- struction, GL, LH have to one another the given ratio of the segments of the base. 60. PROP. LXXXIL XF a parallelogram given in species and magnitude be increased or diminished by a gnomon given in magnitude, the sides of the gnomon are given in magnitude. First, I>et the parallelogram AB given in species and mag- nitude be increased by the given gnomon ECBDFG, each of the straight lines CE, DF is given. Because AB is given in species and magnitude, and that the gnomon ECBDFG is given, therefore the whole space AG is given in magnitude : But AG is also given in species, be- nd cause it is similar ' to AB ; therefore the sides of AG arc ^ j2^. 6. given^ : Each of the straight lines AE, AF ' ^*" is therefore given j and each of the straight yL lines CA, AD is given'', therefore each of » 4.dat. the remainders EC, DF is given^ Next, let the parallelogram AG given in species, and niagnitude, be diminished by the given gnomon ECBDFG, each of the straight lines CE, DF is given. Because the paralleloj^ram AG is given, as «Jcf. H also its gnomon ECBDFG, the remaining space AB is given in magnitude DATA. 433 magnitude : But it is also given in species ; because it is similar' » f |" ^ to AG ; therefore'' its sides CA, AD are given, and each of \ -ii. 6. the straight lines EA, AF is given ; therefore EC, DF are "'^^■d*'- each of them given. The gnomon and its sides CE, DF may be found thus in the first case. Let H be the given space to which the gnomon must be made equal, and find"* a parallelogram similar to AB "25.6. and equal to the figures AB and H together, and place its sides AE, AF from the point A, upon the straight lines AC, AD, and complete the parallelogram AG which is about the same diameter* with AB ; because therefore AG is equal to both *26. 6. AB and H, take away the common part AB, the remaining gnomon ECBDr G is equal to the remaining figure H ; therefore a gnomon equal to H, and its Sides CE, DF are found : And in like manner they may be found in the other case, in which the given figure t^ must be less than the figure FE from which it is to be taken. PROP. LXXXIII. 38, J. F a parallelogram equal to a given space be applied to a given straight line, deficient by a parallelogram given in species; the sides of the defect are given. Let the parallelogram AC equal to a given space be applied to the given straight line AB, deficient by the parallelogram BDCL given in species, each of the straight lines CD, DB are given. Biseft AB in E ; therefore EB is given in'magnitude ; upon EB describe* the parallelogram EF similar to DL and simi- * ^8. 6, larly placed ; therefore EF is given in species, and is about the same diameter'' with DL ; let BCG be the diameter, and constru£l the figure ; therefore, because the figure EF given in species is describ- ed upon the given straight line EB, EF is given= in magnitude, and the gno- A E D B « 56.dat. mon ELH is equal"* to the given figure "3^. and AC i therefore^ since EF is diminished by the given gnomon e 8«.'d»t. ELH, the sides EK, FH of the 2;nomon are given ; but EK is equal to DC, and FH to BD f wherefwc CD, DB are each of them given, F f TJus 434 ' EUCLID'S This demonstration is the analysis of the problem in the 28th Prop of Book 6. the construdiion and demonstration of which proposition is the composition of the analysis ; and be- cause the given space AC, or its equal the gnomon ELH, is to be taken from the figure EF described upon the half of AB similar to BC, therefore AC must not be greater than EF, as is shewn in the 27th Prop. B. 6. 59. PROP. LXXXIV. X F a parallelogram equal to a given space be ap- plied to a given straight line, exceeding by a pa- rallelogram given in species ; the sides of the ex- cess are given. Let the parallelogram AC equal to a given space be applied to the given straight line AB, exceeding by the parallelogram BDCL given in species ; each of the straight lines CD, DB are given. Bisedl AB in E ; therefore EB is given in magnitude: Upon •48. 6. BE describe* the parallelogram EF similarto LD, and similarly placed ; therefore EF is given in species, and is about the * 26. 6, same diameter'' with LD. Let CBG be the diameter, and construct the figure : Therefore, because the figure EF given in species is described upon the given straight lineEB, EFis given in magni- ' 56. dat. tude% and the gnomon ELH is equal * 36. and to the given figure"* ACj wherefore, ■ ■ since EF is increased by the given gnomon ELH, its sides « 82. dat, EK, FH are given-^ j but EK is equal to CD, and FH to BD, therefore CD, DB are each of them given. This demonstration is the analysis of the problem in the aqth Prop. Book 6. the construction and demonstration of which is the composition of the analysis. Cor. if a parallelogram given in species be applied to a given straight line, exceeding by a parallelogram equal to a given space ; the sides of the parallelogram are given. Let the parallelogram ADCE given in species be applied t© the given straight line AB, exceeding by the parallelogram BDCG equal to a given space j the sides A D, DC of the pa- rallelogram are given. Draw DATA. 435 Draw the diameter DE of the parallelogram AGj and con- struct the figure. Because the parallelogram AK is equal* to * -^s. 5. BC which IS given, therefore AK is given ; and BK is similar^ to AC, there- *^^ ^ 5^ '■ -*• 6. fore BK is given in species. And since the parallelogram AK given in magni- rr tude is applied to the given straight line AB, exceeding by the parallelogram BK — given in species, therefore by this pro- *^ position, BD, DK the sides of the excess are given, and the straight line AB is given ; therefore the whole AD, as also DC, to which it has a given ratio is given. PROB. To apply a parallelogram similar to a given one to a given straight line AB, exceeding by a parallelogram equal to a given space. To the given straight line AB apply'^ the parallelogram AK ' 29. 6. equal to the given space, exceeding by the parallelogram i^K similar to the one given. Draw DF, the diameter of BK, and through the point A draw AE parallel to BF, meeting DF produced in E, and complete the parallelogram AC. The parallelogram BC is equal* to AK, that is, to the given space; and the parallelogram AC is similar'' to BK; therefore the parallelogram AC is applied to the straight line AB similar to the one given and exceeding by the parallelo- gram BC which is equal to the given space. PROP. LXXXV. fr*. X F two straight lines contain a parallelogram given in magnitude, in a given angle; if the difference of the straight hnes be given, they shall each of them be given. Let AB, BC contain the parallelogram AC given in magni- tude, in the given angle ABC, and let the excess of BC above AB be given ; each of the straight lines AB, BC is given. Let DC be the given excess of BC above BA, therefore the remainder BD is equal -pt X*- to BA. Complete the parallelogram AD j / / / and because AB is equal to BD, the ratio / / / of AB to BD is given ; and the angle ABD J- - ^ ^ is given, therefore the parallelogram AD is ^ 1> C given in species ; and because the given parallelogram AC is applied to the given straight line DC, exceedmg by the paral- lelogram AD given inspecies, thesides of the excess are given*: • ^- •^'* F f 2 therefore 436 EUCLID'S therefore BD is given ; and DC is given, wherefore the whole BC is given : and AB is given, therefore AB, BC are each of them given. «5 PROP. LXXXVI. XF two straight lines contain a parallelogram given in magnitude, in a given angle ; if both of them to- gether be given, they shall each of them be given. Let the two straight lines AB, BC contain the parallelogram AC given in magnitude, in the given angle ABC, and let AB, BC, together be given ; each of the straight lines AB, BC is given . Produce CB, and make DB equal to BA, and complete the parallelogram ABDE. Because DB is equal to BA, and the angle ABD given, because the adjacent an- gle ABC is given, the parallelogram AD is - J£___A__ given in species : And because AB, BC, to- j 7 7 gether are given, and AB is equal to BD ; / I therefore DC is given: And because the gi- / / / yen parallelogram AC is applied to the given 0^ J3 C straight line DC, deficient by the parallelo- gram AD given in species, the sides AB, BD of the defe£l are » 63. dat. given'jand DC is given, wherefore the remainder BC is given j and each of the straight lines AB, BC is therefore given. 87. PROP. LXXXVII. IF two straight lines contain a parallelogram given in magnitude, in a given angle; if the excess of the square of the greater above the square of the lesser be given, each of the straight lines shall be given. Let the two straight lines AB, BC contain the given paral- lelogram AC in the given angle ABC ; if the excess of the square of BC above the square of B A be given \ AB and BC are each of them given. Let the given excess of the square of BC above the square of BA be the redangle CB, BD ; take this from the square • 2. 2. of BC, the remainder, which is' the re£langle BC, CD is equal to the square of AB : and because the a«gle ABC of the parallelogram AC is given, the ratio of the redangle >«2.iit. of the sides AB, BC to the parallelogram AC is given*" ; and AC is given, therefore the reftangle AB, BC is given j and the redhngle CB, BD is given \ therefore the ratio of the re£l- anglc DATA. 437 «7.dit. angle CB, BD to the rectangle AB, BC, that is% the ratio of \- ^^ the straight line DB to BA is given ; therefore"* the ratio ©f *^'*"'^**' the square of DB to the square of BA is given : And the square of BA is equal to the reftangle BC, CD : wherefore the ra- tio of the redlangle BC, CD to the square of BD is given, as also the ratio of four times the rectangle BC, CD to the square of ED ; and, by composition*, the ratio of four times the rect- angle BC, CD together with the square of BD to the square of BD is given : But four times the redtangle BC, CD, toge- ther with the square of BD, is equal'' to the square of the straight ' 5- 2- lines BC, CD taken together: therefore the ratio of the square of BC, CD, together to the square of BD, is given ; wherefore sche ratio of the straight line BC, together with CD to BD, is given : And, by composition, the ratio of BC together with CD and DB, that is, the ratio of twice BC to BD, is given j therefore the ratio of BC to BD is given, as also"^ the ratio of the square of BC to the rectangle CB, BD : But the re<5langle CB, BD is given, being the given excess of the squa-es of BC, BA; therefore the square of BC, and the straight line BC, is given : And the ratio of BC to BD, as also of BD to BA, has been shewn to be given ; therefore*" the ratio of BC to BA is '' '^'* given ; and BC is given, wherefore BA is given. The preceding demonstration is the analysis of this problem, 58. dat. VIZ. A parallelogram AC, which has a given angle ABC, being given in magnitude, and the excess of the square of BC, one of its sides above the square of the other BA being given ; to find the sides : And the composition is as follows : Let EFG be the given angle to which the angle ABC is required to be equal, and from any point E in F£, draw EG perpendicular to FG ; let the reft- angle EG, GH be the given space to which the parallelogram AC is to be made equal ; and the rectangle HG, GL, be the given excess of the squares of BC, BA. T'~f^ j " A tj~\f Take, in the straight line GE, -t ^ Lr ^ -H.-^ GK equal to FE, and make GM double of GK ; join ML, and in GL produced, take LN equal to LM : Biseft GN in O, »nd between GH, GO find a mean proportional BC : As OG to GL, so make CB to BD ; and make the angle CBA equal F f 3 to 438 EUCLID'S to GFE, and as LG to GK, so make DB to BA, and com- plete the parallelogram AC : AC is equal to the reftangle EG, GH, and the excess of the squares of CB, BA is equal to the reftangle HG, GL. Because as CB to BD, so is OG to GL, the square of CB * ^' ^' is to the re^angle CB, BD as" the reaangle HG, GO to the redlangle HG, GL : and the square of CB is equal to the redlangle HG, GO, because GO, BC, GH, are proportionals ; ' ^*- 5- therefore the reaangle CB, BD is equal" to HG, GL. And because as CB to BD, so is OG to GL ; twice CB is to BD, as twice OG, that is, GN to GL; and, by division, as BC together with Cp is to BD, so is NL, that is, LM, to LG : < 22. 6. Therefore^ the square of BC together with CD is to the square of BD, as the Square of ML to the square of LG : But the *8. 2. square of BC^nd CD together is equal" to four times the re£langle BC, CD together with the square of BD ; therefore four times the r^angle BC, CD together with the square of BD is to the squaYe of BD, as the square of ML to the square of LG : And, by division, four times the redtangle BC, CD is to the square of BD, as the square of MG to the square of GL ; wherefore the rectangle BC, CD is to the square of BD as (the square of KG the half of MG to the square of GL, that is, as) the square of AB to the square of BD, because as LG to GK, so DB was made to BA : Therefore'' the reftan- gle BC, CD is equal to the square of AB. To each of these add the rectangle CB, BD, and the square of BC becomes equal to the square of AB, together with the re6langle CB, BD therefore this redtangie, that is, the given rectangle HG, GL,j is the excess of the squares of BC, AB. From the point Al draw AP perpendicular to BC, and because the angle ABFJ is equal to the angle EFG, the triangle ABP is equiangulaij to EFG : And DB was made to BA, as LG to GK ; therefom as the reaangle CB, BD to CB, BA, so is the redangle HGI B PD C F G L HN GL to HG, GK ; and as the redangle CB, BA to AP, B( so is (the straight line BA to AP, and so is FE or GK E( DATA. 439 EG^ and so is) the rectangle HG, GK to HG, GE ; therefore, ex jequali, as the rectane;le CB, BD to AP, BC, so is the re<ft- angle HG, GL to EG,''GH : And the rectangle CB, BD is equal to HG, GL ; therefore the rectangle AP, BC, that is, the parallelogram AC, is equal to the given rectangle EG, GH. PROP. LXXXVIII. N. J F t\ro straight lines contain a parallelogram given in magnitude, in a given angle; if the sum of the squares of its sides be given, the sides shall each of them be given. Let the two straight lines AB, BC contain the parallelogram ABCD given in magnitude in the given angle ABC, and let the sum ofthe squares of AB,BC be given ; AB,BC are each of them given. First, let ABC be a right angle: and because twice the rect- angle contained by two equal straight lines is equal to both their squares ; but if two straight lines are un- equal, twice the rectangle contained by them less than the sum of their squares, as is evident from the yth Prop. B. 2. Elem. ; therefore twice the given space, to which space the rectangle of which the sides are to be found is equal, must not be greater than the given sum of the squares of the sides ; And if twice that space be equal to the given sum ofthe squares, the sides of the rectangle must necessarily be equal to one another : Therefore in this case describe a square ABCD equal to the given rectanple, and its sides AB, BC are those which were to be found j For the rectangle AC is equal to the given space, and the sum ofthe squares of its sides AB, BC is equal to twice the rectangle AC, that is, by the hypothesis, to the given space to which the suna of the squares was required to be equal. But if twice the given rectangle be not equal to the given sum of the squares of the sides, it must be less than it, as has been shown. Let ABCD be the rectangle, join i\C arid draw BE perpendicular to it, and complete the rectangle AEBF, and describe the circle ABC about the triangle ABC ; AC is its diameter-i : And because the triangle ABC is simi- * C»r.5. ♦. lar^ to AEB, as AC to CB, so is AB to BE ; therefore the " 8. 6. recungle AC, BE is equal to AB, BC i and the rectangle AB, Ff4 ' BC ce B' 'C ^o E U C L I I>' S BC is given, wherefore AC, BE is given : And because the sum of the squares of AB, BC is given, the square of AC which is -47, 1, equal"^ to that sum is given j and AC itself is therefore given ir> magnitude ; Let AC be liicewise given in position, and the * 32. dat. point A J therefore AF is given'' in po- sition : And the rectangle AC, BE is given, as has been shewn, and AC is ' 61 .dat. given, wherefore^ BE is given in mag-» nitude, as also AF which is equal to it ; and AF is also given in position, and fso. dat. the point A is given, wherefore*^ the point F is given, and the straight line ^ — -r^ ls~f * 3i.dat. FB in positions; And the circumfe- ^' ^ H 1^ •"SS. dat, rence ABC is given in position, wherefore'' the point B is given 1 And the points A, C are given ; therefore the straight 5 29.dat. lines AB, BC are given' in position and magnitude. The sides AB, BC of the rectangle maybe found thusj Let the rectangle GH, GK be the given space to which the rect- angle AB, BC is equal; and let GH, GL be the given rect- angle to which the sum of the squares of AB, BC is equal : " 14. 2. Find ^ a square equal to the rectangle GH, GL ; And let its side AC be given in position; upon AC as a diameter describe the semicircle ABC, and as AC to GH, so make GK to AF, and from the point A place AF at right angles to AC : There- 116. 6. fore the rectangle CA, AF is equal ' to GH, GK ; and, by the hypothesis, twice the rectangle GH, GK is less than GH, GL, that is, than the square of AC; wherefore twice the rectangle CA, AF is less than the square of AC, and the rect- angle CA, AF itself less than half the square of AC, that is, than the rectangle contained by the diameter AC and its half; wherefore AF is less than the semidiameter of the circle, and consequently the straight line drawn through the point F pa- rallel to AC must meet the circumference in two points: Let B be either of them, and join AB, BC, and complete the rect- angle ABCD, ABCD is the rectangle which was to be found: p" 34. 1. Draw BE perpendicular to AC ; therefore BE is equal »" to AF, and because the angle ABC in a semicircle is a right an-- * 8. 6. gle, the rectangle AB, BC is equal'' to AC, BE, that is, to the rectangle CA, AF which is equal to the given rectangle ^47. 1. GH, GK: And the squares of AB, BC are together equal*" tO( the square of AC, that is, to the given rectangle GH, GL But if the given angle ABC of the parallelogram AC be not a right angle, in this case, because ABC is a given angle, th«f- ratio of the rectangle contained by the sides AB, BC to the pa- rallelograin DATA. 44» rallelogram AC is given* j and AC is given, therefore the * 62. dau rectangle AB, BC is given : and the sum of the squares of AB, BC is given ; therefore the sides AB, BC are given by the preceding case. The sides AB, BC, and the parallelogram AG, may be found thus : Let EFG be the given angle of the parallelogram, and from any point E in FE draw EG perpendicular to FG : and let the rectangle EG, FH be the given space to which the pa- rallelogram is to be made equal, and let EF . .^ FK be the given rectangle to which the -p= 4-' s una of the squares of the sides is to be equal . / / And, by the preceding case, find the sides _ / J. of a rectangle which is equal to the given O 1^ rectangle EF, FH, and the squares of the sides of which are together equal to the ] given rectangle EF, FK ; therefore, as was shewn in that case, twice the rectangle EF, FH must not be greater than the rectangle EF, FK: let it be so, and let AB, BC be the sides of the rectangle jo'ned in the angle ABC equal to the given angle EFG, c F llGr K and complete the parallelogram ABCD, which will be that which was to be found : Draw AL perpendicular to BC, and because the angle ABL is equal to EFG, the triangle ABL is equiangular to EFG ; and the parallelogram AC, that is, the rectangle AL, BC, is to the rectangle AB, BC as (the straight line AL to AB, that is, as EG to EF, that is, as) the rectan- gle EG, FH to EF, FH ; and, by the construction, the rect- angle AB, BC is equal to EF, FH, therefore the rectangle AL, BC or, its equal, the parallelogram AC, is equal to the given rectangle EG, FH ; and the squares of AB, BC are together equal, by construction, to the given rectangle EF, FK. 442 E U C L I D'S 96. PROP. LXXXIX. XF two straight lines contain a given parallelogram in a given angle, and if the excess of the square of one of them above a given space, has a given ratio to the square of the other ; each of the straight lines shall be given. Let the two straight lines AB, BC contain the given paral- lelogram AC in the given angle ABC, and let the excess of the square of BC above a given space have a given ratio to the square of AB, each of the straight lines AB, BC is given. Because the excess of the square of BC above a given space has a given ratio to the square of BA, let the rectangle CB, BD be the given space; take this from the square of BC, the ■' 2. 2. remainder, to wit, the rectangle* BC, CD has a given ratio to the square of BA : Draw AE perpendicular to BC, and let the square of BF be equal to the rectangle BC, CD, then, because the angle ABC, as also BEA is given, the *4j. dat. triangle ABE is given ''in species, and the ly ratioof7\E to ABisG;iven: Andbecause the ■^/— / ratioof the rectangle BC, CD, that is, of the square of BF to the square of BA, is -ij— f^^ 7^ given, the jatio of the straight line BF to ^^ ^ "^.08. dat, BA is given"= and the ratio of AE to AB is given, wherefore'- " ^* ^"*" the ratio of A£ to BF is given ; as also the ratio of the rcct- ''^^- ^- angle AE, BC, that is% of the parallelogram AC to the rect- angle FB, BC ;and AC is given, wherefore the redtangle FB, BU'is given. The excess of the square of BC above the square of BF, that is, above the redlangle BC, CD, is given, for it is equal=^ to the given redtangleCB. BD ; therefore, be- cause the reftangle contained by the straight lines FB, BC is given, and also the excess of the square of BC above the square '87. <ijt. of BF ; FB, BC,are each of them given*"; and the ratio of FB to BA is given ; therefore AB, BC are given. The Composition is as follows : Let GHK be the given angle to which the angle of the paralk-logram is to be made equal, and from any point G in HG, draw GK perpendicular tQ HK; let GK, HL be the re<5langle 443 DATA. rectangle to which the parallelogram is to be made equal, and let LH, HM be the rect- angle equal to the given space which is to be taken from the square of one of the sides ; and let the ratio of the remainder to the t/ tI '\f- square of the other side be the same with the -*^ ^ *• *-• ratio of the square of the given straight line NH to the square of the given straight line HG. By help of the 87th dat. find two straight lines BC, BF, which contain a rectangle equal to the given rectangle NH, HL, and such that the excess of the square of BC above the square of BF be equal to the given rectangle LH, HM j and join CB, BF in the angle FBC equal to the given angle GHK : And as NH to HG, so ^„ ' ^/ make FB to BA,and complete the paralle- ^ j^^ l) C- logram AC, and draw AE perpendicular to BC : then AC is equal to the rectangle GK, HL ; and if from the square of < BC, the given rectangle LH, HM be taken, the remainder shall have to the square of B A the same ratio which the square of N H has to the square of HG. Because, by the construction, the square of BC is equal to the square of BF together with the rectangle LH, HM ; if from the square of BC there be taken the rectangle LH, HM, there remains the square of BF, which has s to the square of e 22. c. BA the same ratio which the square of NH has to the square of HG, because, as NH to HG, so FB was made to BA : but as HG to GK, so is BA to AE, because the triangle GHK is equiangular to ABE j therefore, ex aequali, as NH to GK, so is FB to AE ; wherefore ^ the rectangle NH, HL is to the rect- h 1. g.. angle GK, HL, as the rectangle FB, BC to AE, BC ; but by the construction the rectangle NH, HL is equal to FB, BC ; therefore* the rectangle GK, HL is equal to the rectangle i 14 5 AE, BC, that is, to the parallelogram AC. The analysis of this problem might have been made as in the 86th Prop, in the Greek, and the composition of it may be made as that which is in Prop. 87th of this edition. 444 E U e L I D'S PROP. XC. o. XF two straight lines contain a given parallelo- gram in a given angle, and if the square of one of them together with the space, vvhich has a given ratio to the square of the other be given, each of the straio'ht lines shall be'ii'iven. lyst the two straight lines AB, BC contain the given pa- rallelogram AC in the given angle ABC, and let the square of BC together with the space which has a given ratio to the square of AB be given, AB, BC are each of them given. Let the square of BD be the space which has the given ratio to the square of AB ; therefore, by the hypothesis, the square of BC together with the square of BD is given. From the point A, draw AE perpendicular to BC ; and because the angles , 43. dst. ABE, BEA are given, the triangle ABE is given* in species ; therefore the ratio of BA to AE is given : And because the ratio of the square of BD to the square of BA is given, the * 58. dat. ratio of the straight line BD to BA is given'' j and the ratio of " 9. dat. BA to AE is given j therefore'^ the ratio of AE to BD is giv*en, as also the ratio of the re£lang\e AE, BC, that is, of the parallelogram AC to the rectangle DB, BC i and AC is given, therefore the redangle DB, BC is given; and the square of 1> aZ B E C GH K L J ,j ,,^^ BC together with the square of BD is given :» therefore'^ be- cause the re6tangle contained by the two straight lines DB, I>C is given, and the sum of their squares is given: The !Xtraight lines DB, BC are each of them given; and the ratio of DB to BA is given ; therefore AB, BC are given. The Compos ilioji is as follows : Let FGH be the given angle to which the angle of the parallelogram is to be made equal, and from any point F in GF draw FH perpendicular to GH ; and let the rectangle FH, GK be that to which the parallelogram is to be made equal; and let the redangle KG, GL be the space to which the square of DATA. «f one of the sides of the parallelogram, together with the space which has' a given ratio to the square of the other side, is to be made equal ; and let this given ratio be the same which the square of the given straight line MG has to the square of GF. By the 88th dat. find two straight lines DB, BC which con- tain a redangle equal to the given redangle MG, GK, and such that the sum of their squares is equal to the given revS- angle KG, GL ; therefore, by the determination of the pro- blem in that proposition, twice the redangle MG, GK must not be greater than the re^an^le KG, GL. Let it be so, and jom the straight lines DB BC^in the an^e D3C equal to the given angle FGH ; and, as MG to GF^, so make DB,to BA, and complete the parallelogram AC : AC is equal to the rea- A/ 445 /rj HE C ^^ K L angle FH, GK ; and the square of BC together with the square of BD, which, by the construetion, has to the square of BA the given ratio which the square of MG has to the square of GF, is equal, by the construction, to the given rectangle KG, GL. Draw AE perpendicular to BC. Because, as DB to BA, so is MG to GF ; and as BA to AE, so GF to FH ; ex aequali, as DB to AE, so is MG to FH i therefore, as the rectangle DB, BC to AE, BC, so is the rectangle MG, GK to FH, GK ; and the rectangle DB, BC is equal to the rectangle MG, GK ; therefore the rectan- gle AE, BC, that is the parallelogram AC, is equal to the rectangle FH, GK. PROP. XCL J.F a straight line drawn within a circle givTn in magnitude cuts off a segment which contains a gi- ven angle; the straight line is given in magnitude. In the circle ABC given in magnitude, let the straight line AC be drawn, cutting off the segment AEC which contains the given angle AEC ; the straight line AC is given in magnitude. Take D the centre of the circle*, join AD, and produce it ■. , to 446 K U C L I D*5 to E and join EC : The angle ACE being * 31. 3. a right'' angle, is given ; and the angle ' *3- dat. ^EC is given i therefore'^ the triangle ACE is given in species, and the ratio of EA to AC is therefore given, and EA is given in magnitude, because the circle is ^ d 5 jgf^ given"* in magnitude j AC is therefore * 2. dat. given'' in magnitude. PROP. XCII. V a straight line given in magnitude be drawn within a circle given in magnitude, it shall cut off a segment containing a given angle. Let the straight line AC given in magnitude be draw^n within the circle ABC given in magnitude j it shall cut off a segment containmg a given angle. Take D the centre of the circle, join AD and produce it to E, and join EC : And because each of the straight lines EA and AC is given, their ratio is given^ : and the angle ACE is a right angle, therefore ^ the triangle ACE is given** in species, and consequently the angle AEC is given. PROP. XCIII. j[ F from any point in the circumference of a circle given in position two straight lines be drawn, meet- ing the circumference and containing a given angle ; if the point in which one of them meets the cir- cumference again be given, the point in which the other meets it is also given. From any point A in the circumference of a circle ABC given in position, let AB, AC be drawn to the circumference making the given angle BAC -, if the point ^,— -^ B be given, the point C is also given. Take D the centre of the circle, and join BD, DC j and because each of the points B, D is given, BD is given* in g\ position i and because the angle BAC is given, the angle BDC is given'', therefor? I because » 1. dat. > 46. dat. 90. » 29, dat. » 'JO. 3. DATA. 447 because the straight line DC is drawn to the given point D in the straight line BDgiven inposition in the given angle BDC, DC is given'^ in position: And the circumference ABC is '■^-°^-- given in position, therefore"^ the point C is given. * '^^ PROP. XCIV. .«!. J-F from a given point, a straight line be drawn touching a circle given in position ; the straight line is given in position and magnitude. Let the straight line AB be drawn from the given point A, touching the circle BC given in position ; AB is given in position and magnitude. Take D the centre of the circle, and join DA, DB : Because each of the points D, A is given, the straight line AD is given* in position and magnitude: And DBA is aright*" angle, wherefore DA is a diameter"^ of f the circle DBA, described about the triangle DBA ; and that circle is there- fore given*^ in position : And the circle \^^ ^ ^ *^-^^'- BC is given in position, therefore the point B is given^ The point A is also given: Therefore e-^^ the straight line AB is given* in position and magnitude. PROP. XCV^. 9^2. IF a straight Hue be drawn from a given point without a circle given in position ; the rectangle contained by the segments betwixt the point and 'the circumference of the circle is eiveii. o Let the straight line ABC be drawn from the given point A without the circle BCD given in po- sition, cutting it in B, C j the rectan- gle BA, AC is given. From the point A, draw* AD ^.. touching the circle J therefore AD is V /ij-A. given'' in position and magnitude : \ y " ^- ^^^ And because AD is given, the square ^ ^ of AD is givenS which is equaW to the rectangle BA *'^ ' 56. dat. Therefore the rectangle BA, AC is given 448 E U C L I D'S 93. ■ PROP. XCVI. If a straight line be drawn through a given point within a circle given in position, the rectangle con- tained by the segments betwixt the point and the circumference of the circle is given. Let the straight line BAG be drawn through the given point A within the circle BCE given in position j the rectangle BA, AC is given. Take D the centre of the circle, join AD, and produce it to the points E, F : Because the points A, D are given, the » 29. dat. straight line AD is given* in position ; and the circle BEC is given in position ; » 28. dat. therefore the points E, F are given''} and B^ the point A is given, therefore EA, AF are each of them givan% and the re£t- *35. 3. angle Ey\, AF is therefore given j and it is equal'^ to the re6l« angle BA, AC, which consequently is given. 9*- PROP. XCVII. xF a straight line be drawn within a circle given in magnitude, cutting off a segment containing a given angle ; if the angle in the segment be bisected by a straight line produced till it meets the cir- cumference, the straight lines which contain the given angle shall both of them together have a gi- ven ratio to the straight line which bisects the an- gle. And the rectangle contained by both these lines together which contain the given angle, and the part of the" bisecting line cut off below the base of the segment, shall be given. Let the straight line BC be drawn within the circle ABC given in magnitude, cutting off a segment containing the gi- ven angle BAC, and let the angle BAC be bisected by the straio-ht line AD ; BA together with AC has a given ratio to AD ; and the rectangle contained by BA and AC together, and the straight line ED cut off from AB below BC the base of the segment is given. Join BD j and because BC is drawn within the circle ABC 2 given DATA. 449 »iven in magnitude cutting off the segment BAC, containing the given angle BAC ; BC is given'" in magnitude : By the * ^i- <»»*• same reason BD is given ; therefore'' the ratio of BC to BD " i- dat, is given : And because the angle BAC is bise^ed by AD, as BA to AC, so is'' BE to EC; and, by permutation, as AB*3.6. to BE, so is AC to CE j wherefore^ as BA and AC together ' 12. 5. to BC, so is AC to CE : And because the angle B AE is equal to EAC, and the angle ACE to' -p,^ ^ ^ ^ '21. ADB, the triangle ACE is equian- gular to the triangle ADB; therefore as AC to CE, so is AD to DB : But as AC toCE, so is BA together with AC to BC; as therefore BA and AC to BC, so is AD to DB : and, by permutation, as BA and AC to AD so is BC to BD ; And the ratio of BC to BD is given, there- for* the ratio of B A together with AC to AD is given. Also the redtangle contained by BA and AC together, and DE is given. Because the triangle BDE is equiangular to the triangle ACE, as BD to DE, so is AC to CE ; and as AC to CE, so is BA and AC to BC; therefore as BA and AC to BC, so is BD to DE ; wherefore the reitangie contained by BA and AC together, and DE, is equal to the re(Sangle CB, BD : But CB, BD is given ; therefore the rei^angle contained by BA and AC together, and DE, is given. Otherwhe^ Produce CA, and make AF equal to AB, and join BF ; and because the angle BAC is double^ of each of the angles *>32. j. BFA, BAD, the angle BFA is equal to BAD ; and the angle BCA is equal to BDA, therefore the triangle FCB is equian- gular to ABD : As therefore FC to CB, so is AD to Y>^ ; and, by permutation, as FC, that is, BA and AC together, to AD, so is CB to BD : And the ratio of CB to BD is given, therefore the ratio of BA and AC to AD is given. And because the angle BFC is equal to the angle DAC, that is, to the angle DBC, and the angle ACB equal to th4 ^angle ADB; the triangle FCB is equiangular to BDE, as "therefore FC to CB, so is BD to DE ; therefore the Fedbngle contained by FC, that is, BA and AC together, and DE is G g equal 450 EUCLID'S equal to the re<ftangle CB, BD which is given, and therefore the redangie contained by BA, AC together, and-DE is given. F. PROP. XCVIII. IF a straight line be drawn within a circle given in magnitude, cutting ofFa segment containing a given angle : If the angle adjacent to the angle in the seg- ment be bisected by a straight line produced till it meet the circumference again, and the base of the segment ; the excess of the straight lines which con^ tain the given angle shall have a given ratio to the segment of the bisecting line which is within the circle ; and the rectangle contained by the same ex- cess, and the segment of the bisecting line betwixt the base produced and the point where it again meets the circumference, shall be given. Let the straight line BC be drawn within the circle ABC given in magnitude, cutting ofFa sejTment containing the given angle BAC, and let the angle CAF adjacent to BAC be bi- seiSed by the straight line DAE, meeting the circumference again in D, and BC the base of the segment produced in E ; the excesss of BA, AC has a given ratio to AB ; and the re<5langle which is contained by the same excess and the straight line EJD is given. Join BD, and through B, draw BG parallel to DE meeting AC produced in G: And because BC cuts off from the circle ABC given in magnitude the seg- ment BAC containing a given an- •*i. dat., gle, BC is therefore given" in mag- nitude : By the same reason BD is given, because the angle BAD is equal to the given angle EAF ; therefore the ratio of BC to B D is given : And because the angle CAE IS equal to EAF, of vyhich CAE ^^r, is equal to the alternate angle AGB, and EAF to the interior and opposite angle ABG j therefore the angle AGB is equal to ABG, and the straight line AB equal to AG j so that GC is the DATA. 451 the excess of BA, AC: And because the angle BGC is equal to GAE, that is to EAF, or the angle BAD ; and that the angle BCG is equal to the opposite interior angle BDA of the quadrilateral BCAD in the circle ; therefore the triangle BGC is eqHiangular to BDA . Therefore as GC to CB, so is AD to DB : and, by permutation, as GC which is the ex- cess of BA, AC to AD, so is BC to BD : And the ratio of CB to BD is given ; therefore the ratio of the excess of BA, AC to AD is given. And because the angle GBC is equal to the alternate angle DEB, and the angle BCG equal to BDE ; the triangle BCJG is equiangular to BDE ; Therefore as GC to CB, so is BD to DE : and consequently the rectangle GC, DE is equal to the rectangle CB, BD which is given, because its sides CB, BD are given ; Therefore the rectangle contained by the ex- cess of BA, AC and the straight line DE is given. PROP. XCIX. J-F from the given point in the diameter of a circle given in position, or in the diameter produced, a straight line be drawn to any point in the circumfe- rence, and from that point a straight line be drawn at right angles to the first, and from the point in which this meets the circumference again, a straight line be drawn parallel to the first ; the point in which this parallel meets the diameter is given; and the rectangle contained by the two parallels is given. In BC the diameter of the circle ABC given in position, or in BC produced, let the given point D be taken, and from D let a straight line DA be drawn to any point A in the circum- ference, and let AE be drawn at right angles to DA, and from the point E where it meets the circumference again let EF be drawn parallel to DA meeting BC in F ; the point F is gi- ven, as also the rectangle AD, EF. Produce EF to the circumference in G, and join AG : because GEA is a right angle, the straight line AG is* the diameter pf the circle ABC ; and BC is also a diameter of it; therefore the point H, where they meet, is the centre of the circle, and consequently H is given : And the point Dis given, wherefore DH is given in magnitude. And because AD is G g 2 parallel 95. » Cor. 5. 4. 4S2 »> 4. 6. E U C L I D'S parallel to FG, and GH equal to HA ; DH is equal'' to HF, and AD equal to GF : And DH is given, therefore HF is givea « 30. dat. (J 95. or 96, dat. J>J in magnitude ; and it is also given in position, and the poinfc H is given, therefore<= the point F is given. And because the straight line EFG is drawn from a given point F without or within the circle ABC given in position^ therefore^ the rectangle EF, FG is given : And GF is equal to AD, wherefore the rectangle AD, EF is given. PROP. C. If from a given point in a straight line given in position, a straight line be drawn to any point in the circumference of a circle given in position ; and from this point a straight line be drawn, making with the first an angle equal to the difference of a right angle, and the angle contained by the straight line given in position, and the straight line ^rhich joins the given point and the centre of the circle ; and from the point in which the second line meets, the circumference again, a third straight line be drawn, making with the second an angle equal to that which the first makes with the second : The point in which this third line meets the straight line given in position is given; as also the rectan- gle contained by the first straight line and the seg- ment of the third betwixt the circumference and the straight hne given in position, is given. Let the straight line CD be drawn from the given point C, in the straight line AB given in position, to the circumference of the circle DEF given in position, of which G is the centre; join CG, and from the point D let DF be drawn, making the angle CDF equal to the diflFerence of a right angle, and the angle BCG, and fronti the point F let FE be drawn, making the ., 453 '26.3. DATA. the angle DFE equal to CDF, meeting AB in H : The point H is given j as also the rectangle CD, FH. Let CD, FH n>eet one another in the point K, from which draw KL perpendicular to DF j and let DC meet the circumference again in M, a\ and let FH meet the same in E, and join MG, GF, GH. Because the angles MDF, DFE are equal to one another, the circum- ferences MF, DE are equal* ; and adding or taking away the common part ME, the circumference DM is equal to EF; therefore the straight line DM is equal to the straight line EF, and the angle GMD to the an- gle'' GFE ; and the angles GxMC, GFH are equal to one another, be- cause they are either the same with the angles GMD, GFE, or adjacent to them : And because the angles KDL, LKD are together equal"^ to a right angle, that is, by the hypo- thesis, to the angles KDL, GCB ; the angle GCB or GCH is equal to the angle (LKD, that is, to the angle) LKF or GKH : There- fore the points C, K, H, G are in the circumference of a cir- cle J and the angle GCK is therefore equal to the angle GHF; and the angle GMC is equal to GFH, and the straight line GM to GF ; therefore'' CG is equal to GH, and CM to HF : *26. i. And because CG is equal to GH, the angle GCH is equal to GHC i but the angle GCH is given : Therefore GHC is gi- ven, and consequently the angle CGH is given ; and CG is given in position, and the point G ; therefore' GH is given in position ; and CB is also given in position, wherefore the *" *** point H is given. And because HF is equal to CM, the rectangle DC, FH is equal to DC, CM : But DC, CM is given^ because the point-^^^ »' ^c. C is given, therefore the rectangle DC, FH is given. "S.l. «32. 1. dat. END OF THE DATA. Gg3 455 NOTES ON EU C LID'S DATA DEFINITION II. X HIS is made more explicit than in the Greek text, to pre- vent a mistake which the author of the second demonstra- tion of the 24th Proposition in the Greek edition has fallen into, of thinking that a ratio is given to which another ratio is shewn to be equal, though this other be not exhibited in given magnitudes. See the Notes on that Proposition, which is the 13th in this edition. Besides, by this definition, as it is now given, some propositions are demonstrated, which in the Greek are not so well done, by help of Prop. 2. DEF. IV. In the Greek text, def. 4. is thus : " Points, lines, space?, ** and angles are said to be given in position which have always ** the same situation ;" but this is imperfectand useless, because there are innumerable cases in which things maybe given ac- cording to this definition, and yettheir position cannot be found; for instance, let the triangle ABC be given in position, and let it be proposed to draw a straight line BD from the angle at B to the opposite side AC, which shall cut off the angle DBC, which shall be the -A. seventh part of the angle ABC; suppose ^y^^ \^ this is done, therefore the straight line ^x'''''____— — — ^V^ BD is invariable in its position, that is t » "" ^ — ^ has always the same situation ; for any other straight line drawn from the point B on either side of BD cuts off an angle greater or lesser than the seventh part of the angle ABC; therefore, according to this definition, the straight line BD is given in position, as also* the point D in »28. dat which it meets the straight line AC which is given in position. But from the things here given, neither the straight line BD nor the point D caa be found by the help of Euclid's Ele- ments ool^, by which every thing in his data is supposed may Gg4 be 456 N O T E S O N be found. This definition is therefore of no use. We have amended it by adding, " and which are either actually exhi- " bited- or can be found ;" for nothing is to be reckoned gi- ven, which cannot be found, or is not actually exhibited. The definition of an angle given by position is taken out of the 4th, and given- more distinctly by itself in the definition marked A. DEF. XI. Xli: XIII. XIV. XV. The nth and 12th are omitted, because they cannot be given in English so as to have any tolerable sense ; and there- fore, wherever the terms defined occur, the words which ex- press their meaning are made use of in their place. The 13th, i4thy 15th are omitted, as being of no use. It is to be pbserved in general of the data in this book, that t)iey are to be understood to be given geometrically, not al- ways arithmetically, that is, they cannot always be exhibited in numbers ; for instance, if the side- of a square be given, *44. dat. the ratio of it.to its diameter is given'' geometrically, but not 3. dat. i^ numbers ; and the diameter is given*^; but though the num- ber of any equal parts in the side be given, for example 10, the number of them in the diahieter cannot be given : And the like holds in many other cases^ .j j^ \y., rO ortJ I'X PROPOSITION I. In this it is shewn that A is to B, as C to D, from this, that A is to C, as B to D, and then by permutation ; but it fol- lows directly without these two steps, from 7. to 5. >,n v.u> . PROP. II. The limitation added at the end of this proposition between the inverted commas is quite necessary, because without it the proposition cannot always be demonstrated : For the author having said*, " because A is given, a magnitude equal to it « 1. def. " can be found* : let this be C ; and because the ratio of A to *>Q. def. " B is given, a ratio which is the same to it can be found V addsj " let it be found, and let it be the ratio of C to a." Now, from the second definition, nothing more follows than that some ratio, suppose the ratio of E to Z, can be found, which is the same with the ratio of A to B ; and when the author supposes that the ratio of C to A, which is also * See Dr. Gregory's ^editba of the Dato, i E U C L I D'S D A T A. 4.57 also the same with the ratio of A to B, can be found, he ne- cessarily supposes that to th,e three magnitudes E, Z, C, a fourth proportional A may be found; but this cannot always be done by the Elements of Euclid ; from which it is plaia Euclid must have understood the proposition under the limita- tion which is now added to his vext. An example will make this clear : Let A be a given angle, and B another angle to which A has a given ratio ; for instance, the ratio of the given straight line E to the given one Z ; then, having found an angle C equal to A, how can the an- gle A be found to which C has the C -p . same ratio that E has to Z ? Cer- tainly no way, until it be shown / \ Z how to find an angle to which a / \ given angle has a given ratio, which cannot be done by Euclid's Elements, nor probably by any Geometry known in his time. Therefore, in all the proposi- tions of this book which depend upon this second, the above- mentioned limitation must be understood, though it be not explicitly mentioned. PROP. V. The order of the Propositions in the Greek text between Prop. 4. and Prop. 25. is now changed into another which is more natural, by placing those which are more simple before those which are more complex ; and by placing together those which are of the same kind, some of which were mixed among others of a different kind. Thus, Prop. 12. in the Greek is now made the 5th, and those which were the 22d and 23d are made the 1 ith and 12th, as they are more simple than the propositions concerning magnitudes, the excess of one of which above a given magnitude has a given ratio to the other, after which these two were placed ; and the 24th in the Greek text is, for the same reason, made the 13th. PROP. VI. VII. These are universally true, though, in the Greek text, they are demonstrated by Prop. 2. which has a limitation ; .they are therefore now shewn without it. 458 N O T E S O N PROP. XIL In the 23d Prop, in the Greek text, which here is the 12th, the words, " /x*! rer auras b»," are wrong translated by Claud. Hardy, in his edition of Euclid's Data, printed at Paris, anno 1625, which was the first edition of the Greek text; and Dr. Gregory follows him in translating them by the words, " etsi, *' non easdem," as if the Greek had been n v.xt (/.-n Im »vrus as in Prop. 9. of the Greek text. Euclid's meanmg is, that the ratios mentioned in the proposition must not be the same ; for, if they were, the proposition would not be true. What- ever ratio the whole has to the whole, if the ratios of the parts of the first to the parts of the other be the same with this ra- tio, one part of the first may be double, triple, &c. of the other part of it, or have any other ratio to it, and consequently cannot have a given ratio to it j wherefore, these words must be rendered by " non autem, easdem," but not the same ratios, as Zambertus has translated them in his edition. PROP. XIII. Some very ignorant editor has given a second demonstration of this proposition in the Greek text, which has been as igno- rantly kept in by Claud. Hardy and Dr. Gregory, and has been retained in the translations of Zambertus and others ; Carolus Renaldinus gives it only : The author of it has thought that a ratio was given, if another ratio could be shewn to be the same to it, though this last ratio be not found : But this is al- together absurd, because from it would be deduced that the ratio of the sides of any two squares is given, and the ratio of the di- ameters of any two circles, &c. And it is to be observed, that the moderns frequently take given ratios, and ratios that are always, the same, for one and the same thing } and Sir Isaac Newton has fallen into this mistake is the 17th Lemma of his Principia, edit. 1713, and in other places i but this chould be carefully avoided, as it may lead into other errors. PROP. XIV. XV. Euclid in this book has several propositions concerning magnitudes, the excess of one of which above a given magni- tude E U C L I D'S D A T A. 459 tude has a given ratio to the other ; but he has given none concerning magnitudes whereof one together with a given magnitude has a given ratio to the other ; thousih thess last occur as frequently in the solution of problems as the first; the reason of which is, that the last may be all demonstrated by help of the first ; for it a magnitude, together with a given magni- tude, has a given ratio to another magnitude, the excess of this other above a given magnitude shall have a given ratio to the first, and on the contrary ; as we have demonstrated in Prop. 14 And for a like reason, Prop. 15. has been added to the Data. One example will make the thing clear : Sup- pose it were to be demonstrated, that if a magnitude A toge- ther with a given magnitude has a given ratio to another mag- nitude B, that the two magnitudes A and B, together with a given magnitude, have a given ratio to that other magnitude B ; which is the same proposition with respect to the last kind of magnitudes above-mentioned, that the first part of Prop. 16, in this edition is in respectof the first kind : This is shewn thus, from the hypothesis, and by the first part of Prop. 14. the excess of B above a given magnitude has unto A a given ratio; and, therefore, by the first part of Prop. 17. the excess of B above a given magnitude has unto B and A together a given ratio ; and by the second part of Prop. 14. A and B to- gether with a given magnitude has unto B a given ratio; which is the thing that was to be demonstrated. In like man- ner, the other propositions concerning the last kind of mag- nitudes may be shewn. PROP. XVI. XVII. In the third part of Prop. i®. in the Greek text, which Is the i6th in this edition, after the ratio of EC to CB has been shown to be given; from this, by inversion and conversion the ratio of BC to BE is demonstrated to be given; but with- out these two steps, the conclusion should have been made only by citing the 6th Proposition. And in like manner, in the first part of Prop. 11. in the Greek, which in this edition is the 17th from the ratio of DB to BC being given, the ratio of DC to DB is shewn to be given, by inversion and composi- tion, instead of citing Prop. 7. and the same fault occurs in the second part of the same Prop. 11. 46(5 N O T E S O N PROP. XXI. XXII. These now are added, as being wanting to complete the subject treated of in the four preceding propositions. PROP. XXIII. This which is Prop. 20. in the Greek text, was separated from Prop. 14. 15. ib. in that text, after which it should have been immediately placed, as being of the same kind ; it is now put into its proper place j but Prop. 21. in the Greek is left out, as being the same with Prop. 14. in that text, which is here Prop. 18. PROP. XXIV. This, which is Prop. 13. in the Greek, is now put into its proper place, having been disjoined from the three follow^ing it in this edition, which are of the same kind, PROP. XXVIII. This, which in the Greek text is Prop. 25. and several of the following propositions, are there deduced from Def. 4. which , is not sufficient, as has been mentioned in the note on that de- finition i They are therefore now shewn more explicitly. ' PROP. XXXIV. XXXVI. Each of these has a determination, which is now addedj which occasions a change in their demonstrations. PROP. XXXVII. XXXIX. XL. XLI. The 35th and 36th Propositions in the Greek text are joined into one, which makes the 39th in this edition, because the same enunciation and demonstration serves both : And for the same reason Prop, 37. 38. in the Greek are joined into one, which here is the 40th. Prop. 37. is added to the Data, as it frequently occurs in the solution of problems j and Prop. 41. is added, to com- plete the rest. PROP. XLII. This is Prop. 39. in the Greek text, whero the whole con- struction of Prop. 22. of Bouk I. of the Elements is put, with- out need, into the demonstration, but is now only cited. PROP. XI.V. This is Prop. 42. in the Greek, where the three straight lines made use of in the construction are said, but not shewn, to be such that any two of them is greater than the third, which is now done. i E U C L I D ' S D A T A. 461 PROP. XLVIT. This is Prop. 44. in the Greek text ; but the demonstration of it is changed into another, wherein the several cases of it are shewn, which, though necessary, is not done in the Greelc. PROP. XLVIII. There are two cases in this proposition, arising from the two cases of the third part of Prop. 47. on which the 48th depends ; and in the composition these two cases are expli- citly given. PROP. LII. The construction and demonstration of this, which is Prop- 48. in the Greek, are made something shorter than in that text, PROP. uir. Prop. 63. in the Greek text is omitted, being only a case of Prop. 49. in that text, which is Prop. 53. in this edition. PROP. LVIII. This is not in the Greek text, but its demonstration is con' tained in that of the first part of Prop. 54. in that text ; which proposition is concerning figures that are given in species: This 58th is true of similar figures, though they be not given in species, and, as it frequently occurs, it was necessary to add itj PROP. LIX. LXI. This is the 54th in the Greek ; and the 77th in the Greek, being the very same with it, is left out, and a shorter demon- stration is given of Prop. 6i. PROP. LXII. This, which is most frequently useful, is not in the Greek^ and is necessary to Prop. 87. 88. in this edition, as also, though not mentioned, to Prop. 86. 87. in the former editions. Prop. 66. in the Greek text is made a corollary to it. PROP. LXIV. This contains both Prop. 74, and 73. in the Greek text; the first case of the 74th is a repetition of Prop. 56. from which it is separated in that text by many propositions ; and as there is no order in these propositions, as they stand in the Greek, they are now put into the ordei which seemed most c<Ml?enient and natural. The 462 N O T E S O N The demonstration of the first part of Prop. 73. in the Greek is grossly vitiated. Dr. Gregory says, that the sentences he has inclosed betwixt two stars are superfluous, and ought to be can- celled ; but he has not observed, that what follows them is ab- surd, being to prove that the ratio [See his figure] of AT to FK is given, which, by th- hypothesis at the beginning of the proposicion, is expressly given j so that the whole of this part was to be altered, which is done in this Prop. 64. PROP. LXVII. LXVIII. Prop. 70, in the Greek text, is divided into these two, for the sake of distindtness ; and the demonstration of the 67th is rendered shorter than that of the first part of Prop. 70. in the Greek, by means of Prop. 23. of Book 6. of the Elements. PROP. LXX. This is Prop. 62. in the Greek text ; Prop. 78. in that text is only a particular case of it, and is therefore omitted. Dr. Gregory, in the demonstration of Prop. 62. cites the 49th Prop. dat. to prove that the ratio of the figure AEB to the parallelogram AH is given; whereas this was shewn a few lines before : And besides, the 49th Prop, is not applicable to these two figures ; because AH is not given in species, but is, by the step for which the citation is brought, proved to be given in species. PROP. LXXill. Prop. 83. in the Greek text, is neither well eniinciated nor demonstrated. The 73d, which in this edition is put in place of it, is really the same, as will appear by considering [See Dr. Gregory's edition], that A, B, F, E, in the'Greek text, are four proportionals, and that the proposition is to shew, that A, which has a given ratio to E, is to F, as B is to a straight line to which A has a given ratio; or, by inversion, that F is to A, as a straight line to which A has a given ratio, is to B : that is, if the proportionals be placed in this order, viz. F, E, A, B, that the first F is to A, to which the second E has a given ratio, as. a straight line to which the third A has a given ratio is to the fourth B; which is the enuncia;tion of this 73d, and was thus changed, that it might be made like to that of Prop. 72. in this edition, which is the 82d in the Greek te^t : And the de- 3 monstratioa EUCLID'SDATA. 463 monstration of Prop. 73. is the same with that of Prop. 72. only making use of Prop. 23. instead of Prop. 22. of Book 5. of the Elements. PROP. LXXVII. This is put in place of Prop. 79. in the Greek text, which is not a datum, but a theorem premised as a lemma to Prop. 80. in that text: And Prop. 79. is made Cor. i. to Prop. 77. in * this edition. CI. Hardy, in his edition of the Data, takes no- tice, that in Prop 80. of the Greek text, the parallel KL in the figure of Prop. 77. in this edition, must meet the circumfe- rence, but does not demonstrate it, which is done here at the end of Cor. 3. Prop. 77. in the construction for finding a tri- angle similar to ABC. PROP. LXXVIII. The demonstration of this, which is Prop. 80. in the Gneek is rendered a good deal shorter by help of Prop. 77. PROP. LXXIX. LXXX. LXXXI. These are added to Euclid's Data, as propositions which are often useful in the solution of problems. PROP. LXXXII. This, which is Prop. 60. in the Greek text, is placed before the 83d and 84th, which in the Greek are the 58th and 59th, because the demonstration of these two in this edition are dedu- ced from that of Prop. 82. from which they naturally follow. PROP. LXXXVIII. XC. Dr. Gregory, in his preface to Euclid's works, which he published at Oxford in 1703, after having told that he had sup- plied the defects of the Greek text of the Data in mnumerabic places from several manuscripts, and corrected Ci. Hardy's translation by Mr. Bernard's, adds, that the 8bth theorem, "or *' proposition," seemed to be remarkably vitiated, but which could not be restored by help of the manuscripts; thta he gives three difterent translations of it in Lat:n, according to which he thinks it may oe read ; the two first have no distinct meaning, and the third, which he say* is the bebt, tnough it cofi:ains 464 N O T E S O N contains a true proposition, which is the 90th in this edition, has no connection in the least with the Greek text. And it is strange that Dr. Gregory did not observe, that, if Prop. 86. was changed into this, the demonstration of the 86th must be cancelled, and another put into its place :^ But the truth is, both the enunciation and the demonstration of Prop. 86. are quite entire and right, only Prop. 87. which is more simple, ought to have been placed before it ; and the deficiency which the * Doctor justly observes to be in this part of Euclid's Data, and which, no doubt, is owing to the carelessness and ignorande of the Greek editors, should have been supplied, not by changing Prop. 86. which is both entire and necessary, but by adding the two propositions, which are the 88th and 90th in this edition. PROP. XCVIII. C. These were communicated tome by two excellent geome- ters, the first of them by the Right Honourable the Earl of Stanhope, and the other by Dr. Matthew Stewart j to which I have added the demonstrations. Though the order of the propositions has been in many places changed from that in former editions, yet this will be of little disadvantage, as the ancient geometers never cite the Data, and the moderns very rarely. XjLS that part of the composition of a problem which is its construction may not be so readily deduced from the ana- lysis by beginners, for their sake the following example is given J in which the derivation of the several parts of the con- struction from the analysis is particularly shewn, that they may be assisted to do the like in other problems. PROBLEM. Having given the magnitude of a parallelogram, the angle of which ABC is given, and also the excess of the square of its- sides BC above the square of the side AB j to find its sides and describe it. The analysis of this is the same with the demonstration of the 87th Prop, of the Data, and the construction that is given of the problem at the end of that proposition is thus derived from the analysis. Let EUCLID'S DATA. Let EFG be equal lo the given angle ABC, and because in the analysis it is said that the ratio of the rectangle AB, BC, to the parallelogram AC is given by the 62d Prop. dat. therefore, from a point in FE, the perpendicular EG is drawn to FG, as the ratio of FE to EG is the ratio of the rectangle 465 rpD c ¥ (t L O HN AH^ BC to the parallelogram AC, by what is shewn at tl-.e end of Prop. 62. Next, the magnitude of AC is exhibited by making the rectangle EG, GH equal to it ; and the given excess of the square of BC above the square of BA, to which excess the rectangle CB, BD is equal, is exhibited by the rectangle HG, GL: Then, in the analysis, the rectangle AB, BC is said to be given, and this is equal to the rectangle FE, GH, because the rectangle AB, BC is to the parallelogram AC, as (FE to EG, that is, as the rectangle) FE, GH to EG, GH ; and the parallelogram AC is equal to the rectangle EG, GH, therefore the rectangle AB, BC, is equal to FE, GH : And consequently the ratio of the rectangle CB, BD, that is, of the rectangle HG, GL, to AB, BC, that is, of the straight line DB to BA, is the same with the ratio (of the rectangle GL, GH to FE, GH, that is) of the straight line GL to FE, which ratio of DB to BA, is the next thing said to be given in the analysis : From this it is plain that the square of FE is to the square of GL, as the square of BA, which is equal to the rectangle BC, CD, is to the square of BD : The ratio of which spaces is the next thing said to be given: And from this it follows, that four times the square of FE is to the square of GL, as four times the rectangle BC, CD is to the square of BD; and, by composition, four times the square of FE, together with the square of GL, is to the square of GL, as four times the rectangle BC, CD, together with the square of BD, is to the square of BD, that is (8. 6.) as the square of the straight lines BC, CD taken together is to the square of BD, which ratio is the next thing said to be given in the analysis : And because four times the; square of FE and the square of GL are to be added together therefore in the perpendicular EG there be taken KG equal to H h FE, 466 N O T E S O N FE, and MG equal to the double of it, because thereby thc- squares of MG, GL, that is, joining ML, the square of ML> is equal to four t,imes the square of FE, and to the square of GL : And because the square of ML is to the square of GL, as the square of the straight line made up of BC and CD is to the square of ^D, therefore (22. 6.) ML is to LG, as BC together with CD is to BD j and, by composition, ML and LG together, that is, producing GL to N, so that ML be equal to LN, the straight line NG is to GL, as twice BC is to BD ; and by taking GO equal to the half of NG, GO is to GL, as BC to BD, the ratio of which is said to be given in the analy-^ sis : And from this it follows, that the rectangle HG, GO is to HG, GL, as the square of BC is to the rectangle CB, BD, ■which is equal to the rectangle HG, GL ; and therefore the square of BC is equal to the rectangle HG, GO; and BC is consequently found by taking a mean proportional betwixt HG and GO, as is said in the construction : And because it was shewn that GO is to GL, as' BC to BD, and that now the three first are found, the fourth BD is found by 12. 6. It was likewise shewn that LG is to FE, or GK, as DB to BA, and the three first are now found, and thereby the fourth BA. Make the angle ABC equal to EFG, and complete the paral- lelogram of which the sides are AB, BC, and the constfuction is finished ; the rest of the composition contains the demon- stration. XXS-the propositions from the J3th to the 28th may ()e thought by beginners to be less useful than the rest, be- cause they cannot so readily see how they are to be made use of in the solution of problems ; on this account the two following problems are added, to shew that they are equally useful with the other propositions, and from which it may be easily judged that many other problems depend upon these propositions. PROBLEM L X O find three straight lines such, that the ratio of the first to the second is given; and if a given straight line be taken from the second, the ratio of the remainder to the third is given ; also the rect- angle contained by the first and third is given. E U C L I D'S D A T A. • 467 Let AB be the first straight line, CD the second, and EF the third : And because the ratio of AB to CD is given, and that if a given straight line be taken from CD, the ratio of the remainder to EF is given ; therefore* the excess of the first AB * 2i. dat. above a given straight line has a given ratio to the third EF : Let BH be that given straight line; therefore AH the excess or AB above it, has a given ratio to EF j and . lt r consequently'* the rectangle BA, AH, has a ^ *? ^ " ^* ^' given ratio to the rectangle AB, EF, which last rectangle is given by the hypothesis; C ' O D thLretorc'= the rectangle BA, AH is given, * «. dau . and BH the excessof its sides is given ; where- K F fore the sides AB, AH are given'' : And be- * 85. dat cause the ratios of AB to CD, and of AH to jC NM^^ Q EF are given, CD and EF are ' given. The Composition. Let the given ratio of KL to KM be that which AB is required to have to CD ; and let DG be the given straight line which is to be taken from CD, and let the given ratio of KM to KN be that which the remainder must have to EF j also let the given rectangle NK, KO be that to which the rectangle AB, EF is required to be equal : Find the given straight line BH which is to be taken from AB, which is done, as plainly appears from Prop. 24. dat. by making as KM to KL, so GD to H B. To the given straight line BH apply' a re£langle equal • t9. 6. to LK, KO exceeding by a square, and let BA, AH be its sides : Then is AB the first of the straight lines required to be found, and by making as LK to KM, so AB to DC, DC will be the second : And lastly, make as KM to KN, so CG to EF, and EF is the third. For as AB to CD, so is HB to GD, each of these ratios being the same with the ratio of LK to KM ; therefore*" AH '19. 5. is to CG, as (AB to CD, that is, as) LK to KM; and as CG to EF, so is KM to KN ; wherefore, ex aequali, as AH to EF, so is LK to KN : And as the rectangle BA, AH to the rectangle BA, EF, so is* the rectangle LK, KO to the ■ i.g. rectangle KN, KO : And by the construction, the rectangle BA, AH is equal to LK, KO: Therefore '' the rectangle AB, » 14. j. EF is equal to the given rectangle NK, KO : And AB has to CD the given ratio of KL to KM ; and from CD the given straight^ine GD being taken, the remainder CG has to EF the given ratio of KM to KN. Q. E. D. Hh2 46S NOTES ON PROB. II. JL O find three straight lines such, that the ratio of the first to the second is given ; and if a given straight line be taken from the second, the ratio of the remainder to the third is given ; also the sum of the squares of the first and third is given. Let AB be the first straight line, BC the second, and BD the third: And because the ratio of AB to BC is given, and that if a given straight line be taken from BC, the ratio of the » 24. dat, remainder to BD is given ; therefore, the excess of the first AB above a given straight line, has a given ratio to the third BD: Let AE be that given straight line, therefore the remainder EB has a given ratio to BD : Let BD be placed at right angles to *44.dat. EB, and join DE; then the triangle EBD is^ given in species; wherefore the angle BED is given: Lei AE which is given in magnitude, be given also in position, as also the point £, and « 32. dat. the straight line ED will be given'^ in position: Join AD, and ^ 47. 1. because the sum of the squares of AB, BD, that is^, the square of AD is given, therefore the straight line AD is given m mag- « 3*. dat. nitudj; and it is also given'^ in position, because from the given point A it is drawn to the straight line ED given in position : Therefore the point D, in which the two straight lines AD, '28.dat, ED, given in position, cut one another, is given*^ : And the «.S3. dat straight line DB, which is at right angles to AB, is givens in position, and AB is given in position, therefore*^ the point B " 29. dat, is given : And the points A, D are given, wherefore *> the straight lines AB, BD arc given : And the ratio of AB to BC ' 2. dat. is given, and therefore' BC is given. The Composition. Let the given ratio of FG to GH be that which AB is required to have to BC,and let HK be the given straight line which is to be taken from BC, and let the ratio which the F G H X remainder is required to have to BD be the given ratio of HQ to LG, and place GL at right angles to FH, and join LF, LH : Next, E U C L I D'S D A T A. 469 Next, as HG is to GF, so make HK to AE ; produce AE to N, so that AN be the straight line to the square of which tb« sum of the squares of AB, BD is required to be equal; and make the angle NED equal to the angle GFL; and from the centre A, at the distance AN, describe a circle, and let its cir- cumference meet LD in D, and draw DB perpendicular to AN, and DM making the aitgie BDM equal to the angle GLH. Lastly, produce BM to C, so that AlC be equal to KH ; then is AB the ftrst, BC the second, and BD the third of the straight lines that were to be found. For the triangles EBD, FGL, as also DBM, LGH being equiangular, as EB, to BD, so is FG to GL ; and as DB to BiVI, so is LG to GH j therefore, cx aequali, as EB to BM, so is (FG to GH, and so is) AE to HK or MC ; wherefore'', * 12.5. AB is to BC, as AE to HK, that is, as FG co GH, that is, in the given ratio : and from the straight line BC taking MC, which is equal to the given straight line HK, the remainder BM has to BD the given ratio of HG to GL : and the sum of the squares of AB, BD is equai^" to the square of AD or "* *' ^ AN, which is the given space. Q. E. D. I believe it would be in vain to try to deduce the precedlne construction from an algebraical solution of the problem. £XD OF NOTES TO THE PATA. Uh3 X i THE ELEMENTS OF PLANE AND SPHERICAL TRIGONOMETRY, LONDON: Priiited for F. Winceave, in the Strand, Successor to Mr. NornsE, 1806. PLANE TRIGONOMETRY 473 LEMMA L Fig. i. i^ET ABC be a re£tilineal angle, if about the point B as a centre, and with any distance BA, a circle be described, meeting BA, BC, the straight lines including the angle ABC in A, C ; the angle ABC will be to four right angles, as the arch AC to the whole circumference. Produce AB till it meet the circle again in F, and throu2;h a 4raw DE perpendicular to AB, meeting the circle in D, E. By 33. 6. Elem. the angle ABC is to a right angle ABD, as the arch AC to the arch AD ; and quadrupling the conse- quents, the angle ABC will be to four right angles, as the arch AC to four time^ the arch AD, or to the whole circum- ference. LExMMA IL Fig. 2. _LiET ABC be a plane rectilineal angle as before : About B as a centre with any two distances BD, BA; tet two circles be described meeting BA, BC, in D, E, A, C ; the arch AC will be to the whole circumference of which it is an arch, as the arch DE is to the whole circumference of which it is an arch. By Lemma i. the arch AC is to the whole circumference of which it is an arch, as the angle ABC is to four right an- gles j and by the same Lemma i. the arch DE is to the whole circumference of which it is an arch, as the angle ABC is to four right angles ; therefore tiie.arch AC is to the whole cir- cumference of which it is an arch, as the arch DE to the whole circumference of which it is an arch. DEFINITIONS. Fig. 3. I. J-iET ABC be a plane reflilineal angle ; if about B as a centre, with BA any distance, a circle ACF be described, meeting BA, BC, in A, C ; the arch AC is called the measure of the anirle ABC. \ II. The circumference of a circle is supposed to be divided into 360 474. PLANE TRIGONOMETRY. 360 equal parts called degrees^ and each degree into 6a equal parts called minutes, and f^ach minute into 60 equal parts calL^d seconds, &c. And as many degrees, minutes, seconds, &c. as are contained in any arch, ot so many de- grees, minutes, seconds, &c. is the angle, of which that arch is the measure, said to be. Cor. Whatever be the radius of the circle of which the mea- sure of a given angle is an arch, that arch will contain the same number of degrees, minutes, seconds, &c. as is mani- fest from Lemma 2. III. Let AB be produced till it meet the circle again in F, the angle CBF, which, together with ABC, is equal to two right angles, is called the «S«^^/^»;^«? of the angle ABC. IV. A straight line CD drawn through C, one of the extremities of the arch AC perpendicular upon the diameter passing through the other extremity A is called the Sine of the arch AC, or of the angle ABC, of which it is the'measure. Cor. The Sine of a quadrant, or of a right angle, is equal to the radius. V. The segment DA of the diameter passing through A, one extremity of the arch AC between the sine CD, and that extremity, is called the Fersed Sine of the arch AC, or an- gle ABC. VI. A straight line AE touching the circle at A, one extremity of the arch AC, and meeting the diameter BC passing through the other extremity C in E, is called the Tangent of the arch AC, or of the angle ABC. VII. The straight line BE between the centre and the extremity of the tangent AE, is called the Secant of the arch AC, or angle ABC. Cor. to def. 4. 6. 7. the sine, tangent, and secant of any an- gle ABC, arc likewise the sine, tangent, and secant of its supplement CBF. It is manifest from def. 4. that CD is the sine of the angle CBF. Let CB be produced till it meet the circle again in Gj and it is manifest that AE is the tangent, and BE the se- cant, of the angle ABG or EBF, from def. 6. 7. '- CoR.I PLANE TRIGONOMETRY. 475; Cor. to def. 4. 5. 6. 7. The sine, versed sine, tangent, and ^'S- * secant, of any arch which is the measure of any given angle ABC, is to the sine, versed sine, tangent, and secant, of any other arch which is the measure of the same angle, as the radius of the first is to the radius of the second. Let AC, MN be me^asures of the angle ABC, according to def. I. CD the sine, DA the versed sine, AE the tangent, and BE the secant of the arch AC, according to def. 4. 5. 6. 7. and N O the sine, OM the versed sine, MP the tangent, and BP the secant of the arch MN, according to the same defini- tions. Since CD, NO, AE, MP are parallel, CD is to NO as the radius CB to the radius NB, and AE to MP as AB to BM, and BC or BA to BD, as BNor BM to BO ; and, by conversion, DA to MO as AB to MB, Hence the corollary is manifest ; therefore, if the radius be supposed to be di- vided into any given number of equal parts, the sine, versed sine, tangent, and secant of any given angle, will each con- tain a given number of these parts; and, by trigonometrical tables, the length of the sine, versed sine, tangent, and se- cant of any angle may be found in parts of which the radius contains a given number ; and, vice versa, a number expres- sing the length of the sine, versed sine, "tangent, and secant being given, the angle of which it is the sine, versed sine, tangent, and secant, may be found. Vin. Fig. a. The difference of an angle from a right angle, is called the complement of that angle. Thus, if BH be drawn perpen- dicular to AB, the angle CBH will be the compliment of the angle ABC, or of CBF. IX. Let HK be the tangent, CL or DB, which iseenjal to It, th* sine, and BK the secant of CBH, the complement "of.^BC, according to def. 4. 6. 7. HK is called the cotangent, BD the cosine^ and BK the ctsecant of the angle ABC. CoR. I, The radius is a mean proportional between the tan- gent and cotangent. For, since HK, BA are parallel, the angles HKB, ABC will be equal, and the angles KHB, BAE are right i therefore the 47^ PLANE TRIGONOMETY. the triangles BAE, KHB are similar, and therefore AE is to AB, asBHorBAtoHK. Cor. 2. The radius is a mean proportional between the co- sine and secant of any angle ABC. Since CD, AE are parallel, BD is to BC or BA, as BA t* BE. PROP. I. Fig. 5. In a right angled plane triangle, if the hypothe- nuse be made radius, the sides become the sines of the angles opposite to them ; and if either side, be ^ made radius, the remaining side is the tangent of '^ the angle opposite to it, and the hypothenuse the secant of the same angle. Let ABC be a right angled triangle ; if the hypothenuse BC be made radius, either of the sides AC will be the sine of the angle ABC opposite to it; and if either side BA be made radius, the other side AC will be the tangent of the angle ABC opposite to it, and the hypothenuse BC the secant of the same angle. About B as a centre, with BC, BA for distances, let two circles CD, EA be described, meeting BA, BC in D, E: Since CAB is a right angle, BC being radius, AC is the sine of the angle ABC, by def. 4. and BA being radius, AC is the tan- gent, and BC the secant of the angle ABC, by def. 6. 7. Cor. I. Of the hypothenuse a side and an angle of a right angled triangle, any two being given, the third is also given. Cor. 2. Of the two sides and an angle of a right angled triangle, any two being given, the third is also given. PROP. n. Fig. 6. 7.^ 1 HE sides of a plane triangle are to one another, as the sines of the angles opposite to them. hi right angled triangles, this Prop, is manifest from Prop. i. for if the hypothenuse be made radius, the sides are the sines of the angles opposite to them, and the radius is the sine of aright angle (cor. to def. 4.), which is opposite to the hypothenuse. In PLANE TRIGONOMETRY. In any oblique angled triangle ABC, any two sides AB, AC will be to one another as the sines of the angles ACB, ABC, which are opposite to them. From C, B draw CE, BD perpendicular upon the opposite sides AB, AC produced, if need be. Since CEB, CE)B are rightangles, BC being radius, CE is the sine of the angle CB A, and BD the sine of the angle ACB ; but the two triangles CAE, DAB have each a right angle at D and E j and like- wise the common angle CAB ; thereKire they are similar, and consequently, CA is to AB, as CE to DB ; that is, the sides are as the sines of the angles opposite to them. Cor. Henre ot two sides, and two angles opposite to them, in a plain triangle, any three being given, the fourth is also given. PROP. III. Fig. 8. -i-N a plane triangle, thesum of any two sides is to their difference, as the tangent of half the sum of the angles at the hase, to the tangent of half their difference. Let ABC be a plane triangle, the sum of any two sides AB, AC will be to their difference as the tangent of half the sum of the angles at the base ABC, ACB to the tangent of half their difference. About A as a centre, with AB the greater side for a distance let a circle be described, meeting AC produced in E, F, and BC in D ; join DA, EB, FB : and draw FG parallel to BC, meeting EB in G. The angle EAB (32. i.) is equal to the sum of the an^rles at the base, and the angle EFB at the circumference is equal to the half of EAB at the centre (20. 3.) ; therefore EFB is half the sum of the angles at the base ; but the antrle ACB (32. I.) is equal to the angles CAD and ADC, or ABC to- gether ; therefore FAD is the difference of the angles at the base, and FBD at the circumference, or BFG, on account of the parallels FG, BD, is the half of that differtnce; but since the angle EBF in a semicircle is a right angle (i. of this), FB being radius, BE, BG are the tangents of the angles EFB, BFG ; but it is manifest that EC is the sum of the sides £A, AC, and CF their difference ; and since BC, FG are parallel (2 6.) EC is to CFj as EB to BG ; that is, the sum of the sides 477 478 PLANE TRIGONOMETRY. sides is to their difference, as the tangent of half the sum of the angles at the base to the tangent of half their difference'. PROP. IV. Fic. 18. xN any plane triangle BAG, whose two sides arc BA, AC, and base BC, the less of the two sides which let be BA, is to the greater AC as the radius is to the tangent of an angle, and the radius is to the tangent of the excess of this angle above half a right angle as the tangent of half the sum of the angles B and C at the base, is to the tangent of half their difference. At the point A draw the straight line EAD perpendicular to BA J make AE, AF, each equal to AB, and AD to AC ; join BE, BF, BD, and from D, draw DG perpendicular upon BF. And because BA is at right angles to EP\ and EA, AB, AF are equal, e«ch of the angles EBA, ABF is half a right angle, and the whole EBF is a right angle (also 4. i. El.) j EB is equal to BF. And since EBF, FGD are right angles, EB is parallel to GD, and the triangles EBF\ FGl3 are simi- lar i therefore EB is to BF, as DG to GF, and EB being equal to BF, FG must be equal to GD. And because BAD is a right angle, BA the less side is to AD or AC the great- er as the radius is to the tangent of the angle ABDj and because BGD is a right angle, BG is to GD or GF as the radius is to the tangent of GBD, which is the excess of thea;ngle ABD above ABF half a right angle. But because EB is parallel to GD, BG is to GF as ED is to DF, that is, since ED is the sum of the sides BA, AC, and FD their dif- ference (3. of this), as the tangent of half the sum of the an- gles B, C, at the base to the tangent of half their difference. Therefore, m any plane triangle, &c. Q. E. D. PROP. V. Fig. 9. and lo. J-N any triangle, twice the rectangle contained by any two sides is to the difference of the sum of the squares of these two sides, and the square of the base, as the radius is to the cosine of the angle in- cluded by the two sides. Let ABC be a plane triangle, twice the rectangle ABD con- tained by any two sides BA, BC is to the difference of the su^ of PLANE TRIGONOMETRY. 475 of the squares of BA, BC, and the square of the base AC, as the radius to the cosine of the angle ABC. From A, draw AD perpendicular upon the opposite side BC, then (by 12. and 13. 2. El.) the difference ot the sum of the squaies of AB, BC, and the square of the base AC, is er^ual to twice the rectangle CBD ; but twice the rectangle CBA is to twice the rectangle CBD ; that is, to the differ- ence of the sum of the squares of AB, BC, and the square of AC (i. 6.) as AB to BD ; that is, by Prop. i. as radius to the sine ot BAD, which is the complement of the angle ABC, that is, as radius to the cosine of ABC. PROP. VI. Fig. IJ. JlN any triangle ABC, whose two sidesare AB, AC, and base BC, the rectangle contained liy half the pe- rimeter, and the excess of it above the base BC, is to the rectangle contained by the straiglit lines by which the half of the perimeter exceeds, the other two sides AB, AC, as the square of the radius is to the square of the tangent of half the angle BAC opposite to the base. Let the angles BAC, ABC be bisected by the straight lines AG, BG i a. id producing the side AB, let the exterior angle CBH be bisected by the straight line BK, meeting AG in K; and from the points G, K, let there be drawn perpendicular upon the sides the straight lines GD, GE, GF, KH, KL, KM, Since therefore (4, 4.) G is the centre « the circle in- scribed in the triangle ABC, GD, GF, GE will be equal, and AD will be equal to AE, BD to BF, and CE to CF. In like manner KH, KL, KM Will be equal, and BH will be equal to.BM and AH to AL, because the angles HBM, HAL are bisected by the straight lines BK, KA : And because in the triangles KCL, KCM, the sides LK, KM are equal, KC is common, and KLC, KMCare right angles, CL will be equal to CM: Since therefore BM is equal to BH,and CM to C'L; BC will be e4ual to BH and CL together j and, adding AB ajid AC together, AB, AC, and BC will together be equal to AH and AL together : But AH, AL are equal : Wherefore each of them is equal to half the perimeter of the triangle ABC : But since AD, AE are equal, and BD, BF, and also CE, CF, AB, together with FC, will be equal to half the perimeter of the triangle to which AH or AL was. shewn to be equal i taking away therefore ibe common AB, the remainder FC 48o PLANE TRIGONOMETRY. FC will be equal to the remainder BH : In the same manner it is demonstrated, that BF is equal to CL : And since the points B, D, G, F, are in a circle, the angle DGF will be equal to th- exterior and opposite angle FBH (22, 3.) j where- fore their halves BGD, IIBK will be equal to one another : The right angled triangles BGD, HBK will therefore be equiangular, and GD will be to BD, as BH to HK, and tho redangle contained byGD, HK will be equal to the rectangle DBH or BFC : But since AH is to HK, as AD to DG, the rc£langle HAD (22. 6.) will be to the rectangle contained by HK, DG, or the redtangle BFC (as the square of AD is to the square of DG, that is) as the square oi the radius t'o the square of the tangent of the angle DAG, that is, the half of 33 AC : But HA is half the perinieter of the triangle ABC, and AD is the excess of the same above HD, that is, above the base BC ; but BF or CL is the excess of HA or AI. above the side AC, and FC, or HB, is the excess of the same HA above the side AB ; therefore the rectangle contained by half the perimeter, and the excess of the same above the base, viz. the redangle HAD, is to the redtangle contained by the straight • lines by which the half of the perimeter exceeds the other two side, that is, the redtangle BFC, as the square of the radius is to the square of the tangent of half the angle BAC opposite to the base. Q. E. D. PROP. VH. Fig. 12. 13. In a plane triangle, the base is to the sum of the sides as the difference of the sides is to the sum or difference of the segments of the base made by the perpendicular upon it from the vertex, according as the square of the greater side is greater or less than the sum of the squares of the lesser side and the base. Let ABC be a plane triangle ; if from A the vertex be drawn a straight line AD, perpendicular upon the base BC, the base BC will be to the sum of the sides BA, AC, as the difference of the same sides is to the sum or difference of the segments CD, BD, according as the square of AC the greater side is - greater or less than the sum of the squares of the lesser side AB, and the base BC. About A as a centre, with AC the greater side for a distance, let a circle be described meeting AB produced in K, F, and CB in G : It is manifest, that FB is the sum, and BE the PLANE TRIGOl^OMETRY. 481 the difference of the sides ; and since AD is perpendicular to GC, GD, CD will be equal ; consequently GB will be equal to the sum or difference of the segments CD, BD, according as the perpendicular AD meets the base produced, or the base ; that is (by Conv. 12. 13. 2.), according as the square of AC is greater or less than the sum of the squares of AB, BC : But (by 55. 3.) the redangle CBG is equal to the redangle EBF; that is ( 16. 6. ) BC is to BF, as BE is to BG ; that is, the base .is to the sum of the sides, as the difference of the sides is to the sum or difference of the segments of the base made by the per- pendicular from the vertex, accordmg as the square of the greater side is greater or less than the stem of the squares of the lesser side and the base. Q. E. D. PROP. VIII. PROB. Fig, 14. Jl HE sum and difference of two magnitudes being given, to find them. Half the given sum added to half the given difference, will be the greater, and half the difference snbtracted fromlialf the sum, will be the less. For let AB be the given sum, AC the greater, and BC the less. Let AD b? half the given sum ; and to AD, DB, which are equal, let DC be added ; then AC will be equal to BD, and DC together ; that is, to BC, and twice DC ; conse- quently twice DC is the difference, and DC half tliat diffe- rence ; but AC the greater is equal so AD, DC ; that is, to half the sum added to half the difference, and BC the less is equal to the excess ot BD, half the sum, above DC half the difference. Q^ E. D. SCHOLIUM. Of the six parts of a plain triangle (the three sides and three angles) any three being given, to find the other three is the business of plane trigonometry ; and the several cases of that problem may be resolved by means of the preceding proposi- tions, as in the two following, with the tables annexed. In these, the solution is expressed by a fourth proportional .to three given lines ; but if the given parts be expressed by num- bers from trigonometrical tables, it may be obtained arithme- tically by the common Rule of Three. NoTF. In the tables the following abbreviations are used: R. is put for the Radius; T. for Tangent; and S, for Sine. Degrees, minutes, seconds, &c. are written in this manner : 30* 25' IS", &c, which signiiie* ao degrees, 25 minutes, 13 seconds, &c. I i 482 PLANE TRIGONOMETRY. SOLUTION of the Cases of Right Angled Triangles. Fig. 15. GENERAL PROPOSITION. XN a liglit angled triangle, of the three sides and three angles, any two heing given besides the right angle, the other three may be found, except when the two acute .angles are given ; in which case the ratios of the sides are only given, being the same with the ratios of the sines of the angles opposite to them. It is manifest from 47. i. that of the two sides and hypo- thenuse, if any two be given, the third may also be found. It is also manifest from 32.. i. that if one of the acute angles of a right angled triangle be given, the other is also given, for it is the complement of the former to a right angle. If two angles of any triangle be given, the third is also given, being the supplement of the two given angles to two right angles. The other cases may be resolved by help of the preceding propositions, as in the following table : G iven. Sought. Two sides, AB,! Theangles AC. iB, C. AB,BC,asideand| Theangles the hypothenuse. IB, C. AB, B, a side and The other an angle. side AC. AB and B, a side and an angle. BC and B, the hypothenuse and an angle. The hypo- thenuse BC. The side AC. AB:AC::R:T,B,of wh ich C is the complement Be : BA : : R : S, C, of which B is the complement R:T,B::BA: AC. S, C : R : : BA : BC. R : S, B : : BC : CA. These five cases are resolved by Prop. i. PLANE TRIGONOMETRY 483 SOLUTION ofthe Cases of Oblique Anqled Triangles. GENERAL PROPOSITION. XN an oblique angled triangle, of the three sides and three angles, any three being given, the other three may be found, except when the three angles are given ; in which case the ratios of the sides are only given, being the same with the ratios of the sines of the angles opposite to them. Given. Sought. A, B, and there- BC, AC, fore C, and the side AB. S, C : S, A : : AB : BC, and also S, C : S, B : : AB :AC.(2.) AB, AC, and B, two sides and an angle opposite to one of them. The angles A and C. AC : AB : : S, B, S, C, (2.) This case admits of two solutions j for C may be greater or less than a quadrant. (Cor. todef. 4.) AB, AC, and A, two sides, and the included angle. The angles B and C. AB + AC:AB-AC::T, C + B T, C— B. { J ) \S ) the sum and difference of the angles C, B, being given, each of them is given. (7.) Otherwise Fig. 18. BA:AC::R:T,ABD, and also R : T,ABD— 45= B + C B— C. :T,— .T,-- :(4.) therefore B and C are ^ivenas before, (7.) Fio. 16. 17 1 2 484 PLANE TRIGONOMETRY. Given, Sought. AB, BC, CA, the three sides A,B,C,the three angles 2ACxCB;ACq+CBq ABq : : R : Co S, C. If ABq -f CBq be greater than ABq. Fig. 16. 2ACxCB:ABq— ACq -CBq::R:Co. S,C. If V Bq be greater than ACq + CBq. Fig. i7. (4.) Otherwise Let AB + B C +AC-2 P. P X P — AB : P — AC X P — BC : : Rq : Tq, \ C, and hence C is known Otherwise Let AD be perpendicular toBC. I. If ABq be less than ACq + CBq. FiG. 16. BC : BA + AC : : BA AC : BD — DC, and BC the sum of BD, DC is given ; therefore each of them is given. (7.) 2. If A Bq be greater than ACq+CBq. F1G.17.BC: BA + AC: :BA— AC:BD + DC ; and BC the difltc- rence of BD, DC is given, therefore each of them is given. (7.) And CA : CD : : R : Co S, C. (i.) and C being found, A and B are found by case 2. or 3 HI G O N O 31 ETJR Y. lA B I) C ^X/ Ti ^%-^7- —JB CONSTRUCTIONS 4«5. OF THE TRIGONOMETRICAL CANON. A Trigonometrical Canon is a Table, which, beginning from one second or one minute, orderly expresses the lengths that every sine, tangent, and secant have, in respect of the radius, which is supposed unity ; and is conceived to be divided into looooooo or more decimal parts. And so the sine, tangent, or secant of an arc, may be had by help of this table ; and, contrariwise, a sine, tangent, or secant, being given, we may find the arc it expresses. Take no- tice, that in the following tract, R signifies the radius, S a sine, Cos. a Cosine, T a tangent, and Cot. a cotangent j also AC q signifies the square of the right line AC ; and the marks or characters +, — , zr, :, : :, and v'j are, seve- rally, used to signify addition, subtraction, equality, pro- portionality, and the extraction of the square root. Again, when a line is drawn over the^um or difference of two quantities, then that sum or difference is to be considered as one quantity. Constructions of the Trigonometrical Can^n. PROP. I. THEOR. 1 HE two sides of any right angled triangle be- ing given, the other side is also given. For (by 47. I.) ACq=ABq + BCqand ACq-BCq= ri£.98. AB q and interchangeably AC q—AB qizBCq. Whence, by the extraction of the square root, there i s given AC = • ABq + BC q; and AB= -/ ACq-BC qi and BC= VACq-ABq. PROP. II. PROB. 1 HE sine DE of the arc BD, and the radius CD Fig. at. being given, to find the cosine DF. Ii3 Th« 486 CONSTRUCTIONS OF THE The radius CD, and the sine DE, being given in the right a ngled triangle CPE, there will be given (by the last Prop.) 'v/CDq-DEq=(CE = ) DF. PROP. III. PROB. '^^ ■ 1 HE sine DE of any arc DB being given, to find DM or BM, the sine of half the arc. DE being given, CE (by the last Prop.) will be given, and accordingly EB, which is the difference between the cosine and radius. Therefore DE, EB, being given, in the right angled triangle DBE, there will be given DB, whose half DM is the sine of the arc DLrrf the arc BD. PROP. IV. PROB. - 1 HE sine BM of the arc BL being given, to find th^ sine, of double, that arc. ?ig. 29. The sine BM being given, there will be given (by Prop. 2.) the cosine CM. But the triangles CBM, DBE, are equi- angular, because the angles at E and M are right angles, and the angle at B common : Wherefore (by 4. 6.) we have CB : CM : : (BD, or) 2 BM : DE. Whence, since the . three first terms of this analogy are given, the fourth also, which is the sineof thearc DB,will be known. Cor. Hence CB : 2 CM ; : BD : 2 DE j that is, the ra- dius is to double the cosine of one half of the arc DB, as the subtense of the arc DB is to the subtense of double that arc. ^AlsoCB: 2CM: :(2BM:2DE: :)BM:DE : :iCB : * CM. Wherefore tlie sind of an arc, and the sine of its dou- ble, being given,' the cosine of the arc itself is given. PROP. V. PROB. Pig. 30. j^ pj£ sines of two arcs, BD, FD, being given, to find FT, the sine of the sum, as Hkewise EL, the sine of their difference. Let the radius CD be drawn, and then CO is the cosine of the arcFD, which accordingly is given, and draw OP through O parallel to DK ; al^o let OM, GE, be ^rawn parallel to CB: Then because the triangles CDK, COP, CHI, FOH, FOP4 TRIGONOMETRICAL CANON. 4S7 FOM, are equiangular ; \a the first place CD : DK : : CO : OP, which, consequently, is known. Also we have CD : CK : : FO : FM j and so, likewise, this will be known. But because F0=EO, then will FM=MG=ON ; and so OP + FM=:FI— sine of the sum of the arcs: And OP— FM; ihat is, OP— ON=EL=:sine of the difference of the arcs; which were to be found. Cor. Because the differences of the arcs BE, BD, BF, are equa , the arc BD is an arithmetical mean Between the arcs BE, BF. PROP. VI. THEOR. JL HE same Things being supposed, the radius is to double the cosine of the mean arc. as the sine of the difference is to the difference of the sines of the extremes. For we have CD : CK : : FO : FM ; whence, by doubling Fig. 3f. the consequents, CD : 2 CK : : FO : (2 FM, or) to FG, which is the difference of the sines, EL, FI. Q^ E. D. Cor. If the arc BD be 60 degrees, the difference of the sines FI, EL, will be equal to the sine FO, of the difference. For, in this case, CK is the sine of 30 degrees ; the double whereof is equal to the radius (by 15. 4.) ; and so, «inceCD =.2 CK, we shall have FO=:FG. And, consequently if the two arcs BE, BF, are equidistant from the arc of 60 de- grees, the difference of the sines will be equal to the sine of the difference FD. Cor. 2. Hence, if the sines of all arcs distant from one another by a given interval, be given, from the beginning of a quadrant to 60 degrees, the other sines may be found by one addition only. For the sine of 61 degrees =the sine of 59 degrees + the sine of i degree; and the sine of 62 degreesz: the sine of 58 degrees + the sine of 2 degrees. Also, the sine of 53 Degreesrzthe sine of 57 degrees + the sine of 3 degrees, and so on. Cor. 3. If the sines of all arcs, from the beginning of a quadrant to any part of a quadrant, distant from each other by a given interval, be given, thence we may find the sines of all arcs to the double of that part. For example, let all the sines to 15 degrees be given ; then, by the preceding ana- logy, all the smes to 30 degrees may be found. For the ra- dius is to double the cosine of 15 degrees, as the sine of i de- Xi4 grce 488 CONSTRUCTIONS OF THE gree is to the difference of thesines of 14 degrees, and i6 de- grees : So, also, is the sine of 3 degrees to the difference be- tween the sines of 12 and 18 degrees ; and so on continually, until you come to the sine of 30 degrees. After the same manner, as the radius is to double the cosine of 30 degrees, or to double the sine of 60 degrees, so is the sine of i degree to the difference of the sines of 29 and 31 de- grees : : sine 2 degrees to the' difference of the sines of 28 and 32 degrees : : sine 3 degrees, to the difference of the sines of 27 and 33 degrees. But, in this case, the radius is to double the cosine of 30 degrees, as i to -v/ 3. For (see the figure for Prop. 15, Book IV. of the Ele- ments) the angle BGC— 60 degrees, as the arc BC, its mea- sure, is a sixth part of the whole circumference ; and the straight line BC=:R. Hence it is evident that the sine of 30 degrees is equal to half the radius ; and therefore, by Prop. 2. the cosine of 30 degrees =: v'R*— _r::-^_A-_> and its double ~v'3R*rr:R X V 3. Consequently, radius is to double the cosine of 30" : : R : R X a/3 : : i : \/ 3- And, accordingly, if the sines of the distances from the arc of 30 degrees, be multiplied by -/ 3, the differences of the sines will be had. So, likewise, may the sines of the minutes in the beginning of the quadrant be found, by having the sine, and cosines of one and two minutes given. For, as the radius is to double the cosine of 2' ; : sine i' ; difference of the sines of I'and 3' : : sine 2 : difference of the sines of o' and 4' ; that is, to the sine of 4'. And so, the sines of the four first minutes being given, we may thereby find the sines of the others to 8' and from thence to 16', and so on. PROP. VII. THEOR.' In smalt arcs, the sines and tangents of the same arcs are nearly to one another, in a ratio of equa- lity. ^ig.31. Pqj.^ betause the Triangles CED, CBG, are equiangular, CE : CB : : ED : BG. But as the point -E approaches B, EB will vanish in respect of the arc BD ; whence CE 2 will TRIGONOMETRICAL CANON. 489 will become nearly equal to CB, and so ED will be also nearly equal to BG. If EB be less than the ^ ^ I 0000000 part of the radius, then the difference between the sine and the , . I I tansient will be also less than the part of the • ^ looooeoo ^ tangent. CoR. Since any arc is less than the tangent, and greater than its sine, and the sine and tangent of a very small arc are nearly equal ; it follows, that the arc will be nearly equal to its sine ; And so, in very small arcs, it will be, as arc is to arc, so is sine to sine. To PRO?. VIII. PROB. find the sine of the arc of one minute. The side of a hexagon inscribed In a circle, that is, the subtense of 60 degrees, is equal to the radius (by Coroll. 15th of the 4.th) ; and so the half of the radius will be the sine of the arc of 30 degrees. Wherefore the sine of the arc of 30 degrees being given, the sine of the arc of 15 degrees may be found (by Prop. 3.) Also the sine of the arc of 15 degrees being given (by the same Prop.)* we may have the sine of 7 degrees 30 minutes. So, likewise, can we find the sine of the half of this, viz. 3 degrees 45 minutes; and so on, until 12 bisections being made, we come to an arc of 52% 44', 03*, 45% whose cosine is nearly equal to the radius ; in which case (as is manifest from Prop. 7.) Arcs are proportional to their sines : and so, as the arc of 52% 44% 03*, 45', is to an arc of one minute, so will the sine before found be to the sine of an arc of one minate, which therefore will be given. And when the sine of one minute is found, then (by Prop, 2. and 4.) the sine and cosine of two minutes will be had. PROP. IX. THEOR, If the angle BAC, being in the peripher}- of a .circle, be bisected by the right hne AJD, and if AC be produced until DE=:AD meets it in E ; then willCEzzAB. In the quadrilateral figure ABDC (by 2a. 3.) the angles B and fig. a«. 490 CONSTRUCTIONS OF THE and DC A are equal to two right anglesrrDCE + DCA (by 13. I.) : whence the angle BrrDCE. But, likewise, thean- . gle E=DAC (by 5. i.)=:DAB, and DCrzDB : Wherefore the triangles BAD and CED are congruous, and so C£ is equal to AB. Q, E. D. Fig. S3. PROP. X. THEOR. Let the arcs AB, BC, CD, DE, EF, &c. be eq^^al ; and let the subtenses of the arcs AB, AC, AD, AE, &c. be drawn ; then will AB : AC : : AC :AB + AD::AD:AC + AE::AE:AD + AF::AF :AE + AG. Let AD be produced to H, AE to I, AF to K, and AG to L, ,so that the triangles ACH, ADI, AEK, AFL, be isosceles ones : Then, because the angle BAD is bisected, we shall have DHzrAB (by the last Prop.) j so likewise EI =:AC, FK=AD, also GL:=AE. But the isosceles triangles ABC, ACH, ADI, AEK, AFL, because of the equal angles at the bases are equiangular : Wherefore it will be, as AB : AC : : AC : (APi=) AB + AD : : AD : (Air:) AC + AE : : AE : (AKzr) AD + AF :: AF: (AL = ) AE + AG. Q. E. D. CoR. I. Because AB is to AC, as radius is to <louble the cosine of | the arc AB, (by Coroll. Prop. 4.) it will also be as radius is to double the cosine of | the arc AB, so is 4 AB : i AC : : ^ AC : I AB + i AD : : ^ AD : \ AC + i AE : : I AE : i AD + i AF,&c. Nowlet each of the arcs AB, BC, CD, &c. be 2' ; then will \ AB be the sine of one minute, \ AC the sine of 2 minutes, f AD the sine of 3 minutes, \ AE the sine of 4 minutes, &c. Whence, if the sines of one and two minutes be given, we may easily find all the other sines \vt the following manner. ■ , - Let the cosine of the arc of one minute, that is, the sine of the arc of 89 deg. 59' be called Q_; and make the following analogies ; R.:%Qj.: Sin. 2' : S. i' +S. 3'. Wherefore thai sine of 3 minutes will be given. Also, R. : 2 Q^: : S. 3' :! S. 2' +S. 4'. Wherefore the S. 4' is given. And R. : 2 Qj : : S. 4' : S. 3' -j-S. 5' j and so the sine of 5' will be had. Likewise, R. : 2 Q^: : S. 5' : S 4' +S. 6' ; and so we shall have the sine of 6'. And in like manner, the sines of every minute of the quadrant will be given. And because the ra- dius, or the first term of the analogy, is unity, the operations wiU TRIGONOMETRICAL CANON. will be with great ease and expedition calculated by multipli- cation, and contracted by addition. When the sines are found to 60 degrees, all the other sines may be had by addi- tion only, by Cor. i. Prop. 6. , The sines being given, the tangents and secants may be found from the following analogies (see Figure 3, for the de- finitions) J because the triangles BDC, BAE, BHK are equi- angular, we have BD : DC : : BA : AE ; that is, Cos. : S. : : R. : T. AE : BA : : BH : HK ; that is, T. : R. : : R. : Cot. BD : BC : : BA : BE J that is, Cos. : R. : : R. : Secant, CD : BC : : BH : BKi that is, S. : R. : : R. : Cosec. 49 » 492 OF LOGARITHMS. I. X HE indices or exponents of a series of numbers in geo- metrical progression, proceeding from i, are also called the logarithms of the numbers in that series.* Thus if <2 denote any number, and th^ geometrical scries, i, «*, c% a^^ a\ Sec. be produced by actual multiplication, then i, 2, 3, 4, &5. are called the logarithms of the first, sepond, third, and fourth powers of a respectively. Consequently, if, in the above, a be equal to the number 2, then i is the logarithm of 2, 2 is the logarithm of 4, 3 is the logarithm of 8, 4 is the logarithm of 16, &CC. But if fl be equal to lO, then i is the logarithm of 10, 2 is the logarithm of 100, 3 is the logarithm of lOOO, 4 is the logarithm of lOOOO, &c. The series may be conti- nued both ways from i. Thus — , -> —1 — » i, a, a\ a' /2*, &c. constitute a series in geometrical progression, and, agreeable to the established notation in algebra, the indices, or logarithms are —4, — 3, — 2, — 1,0, 1,2, 3, 4, &c. If^be equal to the number 2, then— 4 is the logarithm of ->, — 3 is the logarithm of— , — 2 is the logarithmof — , — i is the loga- j 8 4 rithmof — , o is the logarithmof i, i is the logarithm of2,&c. If a be equal to 10, then —4 is the logarithm of , —3 ^ ' ^ ° loooo ^ is the logarithm of , — 2 is the logarithm of — , — i is the ^ , 1000 100 logarithm of — , o is the logarithm of i, and i is the loga- rithm of 10, &c. 2. From the above it is evident, that the logarithms of a series of numbers in geometrical progression, constitute a series of numbers in arithmetical progression. Beginning with I, and proceeding towards the right hand, the terms in the geometrical series are produced by multiplication, but their corresponding logarithms are produced by addition. — On the contrary, beginning with i, and proceeding towards the left hand, the terms in the geometrical progression are produced by division, but their corresponding logarithms are produced by subtraction. The • The reader ought to be acquainted with arithmetical and geometrical progressioai and the binomial theorem, before he enters on a perusal of any account of logarithms. OF LOGARITHMS. 4^j 3. The same observations apply to logarithms when they • :e fractions. Thus if a" denote any number, -^ — , — , L 1 L ± aT. a~. a-a . p I, «", a", ««j a", &c. constitute a series of numbers ' geometrical progression, of which — — , — — , — ^ •—- ^> -^ — , i, -t, &c. are the logarithms ; and it is evident that e assertions in the last article hold true both with respect the numbers in geometrical progression and their corres- iding logarithms. As a and n may be taken at pleasure, follows that numbers in very different geometrical pro- essions may have the same logarithms ; and that the same ies of numbers in geometrical progression may have diiFe- it scries of logarithms corresponding to them. 4. If a be an indefinitely small decimal fraction, and suc- sive powers of i -f ^ be raised, then the excess of any power r 4- ^ above that immediately preceding it will be indefi- ely small. Thus let /z = -00000000001, and then i+V^z: JOOOO00000200000000001 ; and i + a^'m -00000000003000- 30000300000000001 ; and proceeding by actual multiplica- n to obtain higher powers of 1 -0000000000 1, it will be md that the dift'erence between two successive powers is ■y small. If instead of supposing, as above, that a-=z ooooooooi, we suppose it only one millionth part of this ue, then the successive powers of i+ff will differ from i another by much smaller decimal fractions. 5. If therefore a be indefinitely small, and successive ;vers of i +df be raised, a series of numbers in geometrical gression will be produced, of which the common numbers 3, 4, 5, &c. will become terms. For on every multipli* ion by I +ff, an indefinitely small addition is made to the /ver multiplied, and by this indefinitely, small addition, the <t higher power is produced. Some power of i + <z will irefore be equal to the number 2, or so nearly equal to it .t they may be considered as equali Continuing tjie ad- ■ icement of the powers of i-\-a, the numbers 3, 4, 5, &c. the same reasons, will fall into the series. 6. The sum of the logarithms of any two numbers is equal the logarithm of the product of the same two numbers, lus if I + a raised to the «"* power be equal to the number ^ and \i i-\-a raised to the rrf^ power be equal to the num- ber 1 494- OF LOGARITHMS. ber M, then, by the preceding articles, n is the logarithm of I +«^" Or of its equal N, and for the same reason m is the logarithm of M. Hence it follows that n-\-m:=zxhc logarithm ofNxM, for NxM=7+^''xi"+^''^=iT^"+'^by the nature of indices. If the logarithm of N be subtracted . from the logarithm of M, the difference is equal to the loga- rithm of the quotient which arises from the division of M by JN. ror-TT — ==— m+^' , by the nature of indices. — vThe addition of logarithms, therefore, answers to the multi- plication of the natural numbers to which they belong; and the subtraction of logarithms answers to the division by the natural numbers to which they belong. 7. If the logarithms of a series of natural numbers be all mul- tiplied by the same number, the several products will have the last-mentioned properties of logarithms. Thus if the indices of all the powers of i -f o be multiplied by /, then, using the no-j tation stated in the last article, the logarithm of N is nl^ and! the logarithm of M is ot/, and the logarithm of N x M is «/+ ? ml; forNxM=iT^"'xFT^'"'=i'+^"'+''", by the n: M M i ture of indices. Also ml— nl— the Iogarithm,of -5— , for-r-rrzj ' ^ „ i — 1~| a^"^^~^- Hence the products arising from the i-\-a)" _ - multiplication of / into the indices of the powers of i +(?, are termed logarithms, as are also all .numbers, which have the properties stated at the end of article 6. it is on account of these properties that logarithms are so very useful in calcula- tions of the highest importance. 8. If the indices of the powers of'i -f^ be multipled by a, the products are called the hyperbolic logarithms of the num- bers equal to the powers of i+a. Thus if the number N be equal to i"+^"j then na is the hyperbolic logarithm of N and if the number M be.equal to i +<3 '", then ffia is the hy- perbolic logarithm of M. Hyperbolic logarithms are not those in common use, but they can be calculated with less laboui than any other kind, and common logarithms are obtainec from them. 9. If successive powers of a very small fraction be raiset they will successively be less and less in value. This trutf OE LOGARITHMS, appears most evident by putt ing the. value in the form of a vulgar fraction. Thus — ] = ; \ = ** IOOOOO| lOOOOOOOOOO lOOOOO) . cCC. I OOOOOOOOOOOOOOO lo. Let it be required to determine the hyperbolic loga- rithm L, of any number N. Using the same notation, as in the preceding articles, l+tfl"=:N, and, by extracting the n* T root of each side of the equation, i + a— N". Put ot=:— , J_ m n and i+A"=N, and then N" zzi+x^ = (by the binomial m—i m—i m—2 theorem) i-\-mx-\-mx X* +»zx x x x^. 2 2 3 .-f Sec. =1 +tf. Now as ff is indefinitely small, the power of i-f ^, which is equal to the number N, must be indefinitely high ; or, which is the same thing, n must be indefinitely great. Consequently m must be indefinitely small, and there- fore may be rejected from the expressions m — i, w— 2, »i— 3, &c. Hence i being taken from each side of the above equa- , mx* mx^ mx* mx^ tion, we have a—mx— \- 1- — &c. Each side ^ ^ + 5 ^ ;,* ^t of this equation being divided by ot, we have —zzx 1 — m 23 \-~ — &c. But m——, and therefore— =:tf«—.r 1- 45 n m 2 Jt' JT* JT* — '"7'"" °^^- — ^ ^^ hyperbolic logarithm of N, bjr article 8. This series, however, if .r be a whole number, does not converge. Let M be a whole number and M= , and then x is 1 ess than I. For multiplying both sides of the equation by I— X- we haveM— Mar=:i, and therefore i—rr=r.r. Now M . , I r . t M =: =:i-}-a>;). Then we have i+ar=^=:^i = * -* I —x\ . — 5- I . > Y \ I— .r' ^=: I --.r^'^(by putting rr: ) = !— rr+rX X P 2 r — I r — 2 *' — r X X X .r' -^ Sec. But for the same reasons 2 3 .as 495 496 OF LOGARITHMS. as above, r must be indefinitely small, and therefore may be rejected from the factors r— i, r— 2, r— 3, &c. Conse- quently, taking i from each side of the above equa^tion, ^rr r.r* rx^ rx'' ^x^ » t> ^ — rx — &c. But— r=: — , andthere^r _ . 2 3 4 5 P J fore dividing the left-hand side of the equation by — , and A-i* futS ly/* «5 the other by — r, we have ep=zx-{- -1— -f ^_ +-1- -{■- {- ^345 &c. r: the hyperbolic logarithm of M. 11. As by the last article, the hyperbolic logarithm of N jr* x^ X* x^ x^ x'' or i+x is X + 1 ;r- + &c. andasthc I X^ X^ X* x^ hyperbolic logarithm ofMor — — is .r + -^ — I 1 -| I X 2345 X x'^ I "^* X -?--{ V &c. the hyperbolic logarithm of N x M, or is equal to the feum of these tw^o series, that is, equal to 2jr + 2.i'' 2x^ 2.f' 1 \-— — h &c. This series converges faster than either df the preceding, and its value may be expressed thus : 2X.r + — +— + 1- &c. / Z S 1 7/ -f I 2« + 2 12. The logarithm of =2 X logarithm of ; 1- ° n ° 2« + 1 i — ^* the logarithm of ^^=:;-^ — ■• For as the addition of loga- 2?7+i\ —^ rithms ansvv^ers to the multiplication of the numbers to which they belong, the logarithm of the square of any number, ts the logarithm of the number multiplied by 2. . Hence the lo . in 4- 2^1* 2n + 2 2«4-2V garithrn of is 2 X logarithm of — ;. But — — — 2«Tl\ 2'^— I 2«TT^^ _2« + 2^* __ 4«'-i-8?? + 4 __ «^+2» + I __ ^2"^l^*-I~2«-f?*-l" 4«' + 4« " «* + « ~ n+1 x«+i_ n+i « X « + I '^ From the preceding articles hyperbqlic logarithms may be calculated, as in the following examples. Exaropl* OF LOGARITHMS. 497 Example i. Required the hyperbolic logarithm of 2, Put »+r , , 2« + 2 4 , 2 n+iV ~2 and then nm, =— , and .. « ' 2«+i 3' 2n+iY-i — rr. I order to proceed by the series in article 11, let - — —=-", and then .m— . Consequently I-T 3 7 r = 0.14285714286 — =0.00097181730 — =0.00001189980 — =0.00000017347 — =0.00000000275 — =0.00000000004 bum of the above terms, • 0.14384103622 2 ^ - I + T 2n+2 4 o ,0 Log. of or or — - 0.28768207244. ** l—X 2W+I 3 / / *M- The double of which is 0.57 5 364 14488, and answers to the first part of the expression in article 12. secondly, let ~=^ ^^^ then 8 + 8.r =9—9.?', and !s I -^ .2' O :iow is equal to — . Consequently, X =0.05882352941 x^ ' — =0.00006784721 3 ' x^ — =0.00000014086 x'' — =0.00000000035 Sum of the above terms 0.05889151783 "* J^Tiv^' ""' ^ 0.H778303566, ^^ which 4^8 OF LOGARITHMS. which answers to the second part of the expression in article 12. Consequently the hyperbolic logarithm of the number 2150.57536414488 + 0.1177830356=0.69314718054. The hyperbolic logarithm of 2 being thus found, that of 4, 8, ifc, and all the other powers of 2 may be obtained by mutiplying the logarithm of 2 by 2, 3, 4, &c. respectively, as is evident from the properties of logarithms stated in ar- ticle "6. Thus, br multiplication, the hyperbolic logarithm of 4 = 1 . 38629436 1 q8 of 8=2.07944154162 &c. ' ■;. From the above the logarithm of 3 may easily be obtained. For 4-^^=4 X — = 3; and therefore as the logarithm of — 3 4 3 was determined above, and also the logarithm of 4. From the logarithm of 4, viz. - i. 38629436108, Subtract the logarithm of -, viz. 0.28768207244, And the logarithm of 3 is - - - - 1.09861228864. Having found the logarithms of 2 and 3, we can find, by addition only^ the' logarithms of all the powers of 2 and 3, and also the logarithms of all the numbers w^hich can be pro- duced by multiplication from 2 and 3. Thus, To the logarithm of 3, viz. . - - i. 09861228864 ^ Add the logarithrh of 2, viz. - - - 0.693147 18054 :. -yi:r. ,ii^; ; ^ ', .;.:*( ' r--^^ And the sum is the logarithm of 6 - 1.79175946918. To this last found add the logarithm of 2, and the sum 2.48490664972 is the logarithm of 12. The hyaerbolic loganthips of other prime numbers may be more readily calculated by attending to the following article. 13. Let a, bj c be three numbers in arithmetical progres- sion, whose common difference is i. Let b be the prime number, whose logarithm is soirght, and a and c even num- bers whose logarithms are known, or easily obtained from others already computed. Then, a being tke least of the three, and the common difference being i, ^7=^—1, and c=^+i. Consequently a x c •=.b—ixb + i—b^—ij and <r<r+ I =^* J and therefore — = . This is a general ex- ac ac pression ^ OF LOGARITHMS. 49f prcssion for the fraction, which it will be proper to put = —y that the series expressing the hyperbolic logarithm may converge quickly. For as = , ac + a:x =. ac I —x ac + i—ac.v—x, and therefore 2acx-\-xzzi^ and : = . Example 2. Required the hyperbolic logarithm of 5. Here a-=.±. rrr6j and .r— ~ — . Consequently, iac-\-i 49 ^ -" X =0.0204081632 — =0.0000028332 — =0.0000000007 Sum of the above terms, 0.0204109971 2 Log of — or— 0.0408219942 I — -r 24 2C • ' But -^x8x3r=25, and the addition of Logarithms an- swers to the multiplication of the natural numbers to which they belong. Consequently, 2^ To the log. of -:^ - - - - 0.0408219942 Add the log. of 8 - - - - -2.0794415422 And also the log. of 3 - - - - i. 0986122890 And the sum is the log. of 25 - 3.2188758254 Thehalf of this, viz. 1.6094379127, is the hyperbolic loga- rithm of 5 i for 5 X 5 = 25. Example 3. Required the hyperbolic logarithm of 7. licre^=r6 <: = ?, and .r= ■— — rr — , and = = 2 tf c 4- 1 97 as I —X — T-. Consequently, 45 Kka .r= 500 ^ OF LOGARITHMS. jr =0.01030927835 —=0.00000036522 =0.00000000002 5 Sum of the terms r - - - .01030964359 2 Log. of -^ - - - - - -0.02061928718 To which add log. of 8 - - - 2.07944154162 And also log. of 6 - - - - 1.79175946918 The sum is the log. of 49 - - 3.89182029798 For ^x6x 8=49. Consequently the half of this, viz. 48 1.94591014899, is the hyperbolic logarithm of'7 j for 7 x 7 If the reader perfectly understand the investigations and examples, already given, h'e will find no difficulty in calcu- lating the hyperbolic logarithms of kigher prime numbers. It will only be necessary for him, in order to guard against any embarrassment, to compute them as they advance in succession above those already mentioned. Thus, after what has been done, it would be proper, first of all, to calcu-. late the hyperbolic logarithm of 1 1, then that of 13, &c. Proceeding according to the method already explained, it will be found that The hyperbolic logarithm of II is 2.3978952^3016 of 1 3 is 2.564999357538 of 17 IS 2.833213344878 • of 19 is 2.944438979941 Logarithms were invented by Lord Neper, Baron of Mer- chiston, in Scotland. In the year 1614 he published at Edin- burgh a small quarto, containing tables of them, of the hyper- bolic kind, and an account of their construction anci use. The discovery afforded the highest pleasure to mathematicians, as they were fully sensible of the very great utility of logarithms j but it was soon suggested by Mr. Briggs, afterwards Saviliari Professor of Geometry in Oxford, that another kind of lo- garithms would be more convenient, for general purposes, than OF LOGARITHMS. 501 than the hyperbolic. That one set of logarithms may be ob- tained from another will readily appear from the following aiiicle. 14. It appears from articles i, 3, and 7, that if all the lo- garith ms ot the geometrical progression i, i+i?'', T+o}', i+fll\ i + a*, I -hfl\% &c. be multiplied or divided by any given lumber, the products aod also the quotients will like- wise be logarithms, for their addition or subtraction will an- swer to the multiplication or division of the terms in the geometrical progression to which they belong. The same terms in the geometrical progression may therefore be repre- sented with different sets or kinds of logarithms in the fol- lowinir manner. l,i-\-a^\ i+« ) i+a^S i-\-a^, i+a>\ i-{-a\% &c. J, rMS i+^H l+«^^i iT^y T+3^7+tfl^ &c. I, f+P'^ ! + '»''", i-i-a^'"i TTa^"^ i + fl'"*, T+tfi^j &c. In these expressions / and m denote any numbers, whole or fractional ; and the positive value of the term in the geo- metrical progression, mider the same number in the index, is understood to be the same in each of the three series. Thus if I +fll* be equal to 7, then i + a ^, is equal to 7, as is also 4 _ i+ai«. If I +a'^ be equal to 10, then i+a'^is equal to 10, 6 as is also i +^ *, Sec. If therefore /, 2 /, 3/, &c. be hyperbolic logarithms, calculated by the methods already explained, the logarithms expressed by — , — ^, -J^, &c. may be derived m fn 7n from them v for the hyperbolic logarithm of any given num- ber is to the logarithm in the last-mentioned set, of the same number, in a given ratio. Thus 4/ : — - : : i : -—- zz-j— s also 6/ : — : : i : -— -=-r~) ^^* m b Im Im 15. Mr. Briggs's suggestion, above alluded to, was that i should be put for the logarithm of 10, and consequently 2 for the logarithm of 100, 3 for the logarithm of 1000, &c. This proposed alteration appears to have met with the full appro- bation of Lord Neper ; and Mr. Briggs afterwards, within- credible labour and perseverance, calculated extensive tables K k 3 of ^2 OF LOGARITHMS. of logarithms of this new kind, which are naw called com- mon logarithms. If the expeditious methods for calculating hyperbolic logarithms, explained in the foregoing articles,* had been known to Mr. Briggs, his trouble would have been comparatively trivial with that which he must have expe- rienced in his operations. 1 6. It has been already determined that the hyperbolic lo- garithm of5 is i,6o94379i«7, and that of 2 is 0.69314718054, and therefore the sum of these logarithms, viz. 2.30258509324 is the hyperbolic logarithm of 10, If, therefore, for the sake of illustration, as in article 14, we suppose i~+^^— 10, and allow, in addition to the hypothesis there formed, that — , m 2 "i A. — , -^, — , &c. denote common logarithms, then 6 Izz m m m 2.30258509324, and — — 1 ; and the ratio for reducing the hyperbolic logarithm of any number to the common logarithm of the same number, is that of 2.30258509324 to i. Thus in order to find the common logarithm of 2, 2.39258509324 : I : : 0.69314718054': 0.3010299956, the common logarithm of 2. The common logarithms of 10 and 2 being known, "We obtain the common logarithm of 5, by subtracting the common logarithm of 2 from i, the common logarithm of to ; for 10 being divided by 2, the quotient is 5. Hence the common logarithm of 5 is 0.6989700044. Again, to find the common logarithm of 3, 2.30258509324 : i : : 1.09861228864 : .4771212546, the common logarithm *)f3. 17. As the constant ratio, for the reduction of hyperbolic 10 common logarithms, is that of 2.30258509324 to i, it is evident that the reduction may be made by multiplying the hyperbolic logarithm, of the number whose common logarithm is sought, by =.4'?42q448i8. 2.30258509324 Thus 1. 945910 14899, the hyperbolic logarithm of 7, being multiplied by .4342944818, the product, viz. .8450980378, &c. is the common logarithm of 7. I'he common logarithms of prime numbers being derived from the hyperbolic, the common logarithms of other num- bers • Some of the principal particulars of tlie foregoing method* were discovered by the celebrated Ti)om8s Simpsen. Se« als» Mr. Hellins' M»lh«maiical Essays, pub- lishedlQ 1788. OE LOGARITHMS. 503 bers may be obtained from those so derived, merely by addi- tion or subtraction. For addition of logarithms, in any set or kind, answers to the multiplication of the natural num- bers to which they belong, and consequent